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\\frac{1}{2}$ M1\n- Finds $y$-coordinates: $(2, 0)$ and $(\\frac{1}{2}, -\\frac{27}{16})$ A1 A1\n- Uses second derivative or gradient testing to determine nature M1\n- Identifies $(\\frac{1}{2}, -\\frac{27}{16})$ as a Minimum A1\n- Identifies $(2, 0)$ as a Point of Inflexion A1","marks":8,"order":0,"markingRubric":{"totalMarks":8,"points":[{"id":"p1","marks":1,"description":"Differentiates the function using the product rule.","code":"M"},{"id":"p2","marks":1,"description":"Factors the derivative expression correctly.","code":"M"},{"id":"p3","marks":1,"description":"Sets the derivative to zero and finds both critical values of x.","code":"M"},{"id":"p4","marks":1,"description":"Correctly finds the y-coordinate for the first stationary point.","code":"A"},{"id":"p5","marks":1,"description":"Correctly finds the y-coordinate for the second stationary point.","code":"A"},{"id":"p6","marks":1,"description":"Employs a valid method (second derivative or gradient test) to test the nature of the points.","code":"M"},{"id":"p7","marks":1,"description":"Identifies the stationary point at x = 0.5 as a Minimum.","code":"A"},{"id":"p8","marks":1,"description":"Identifies the stationary point at x = 2 as a Point of Inflexion.","code":"A"}],"finalAnswers":["(2, 0) and (1/2, -27/16)"],"generatedAt":"2026-05-06T14:04:21.295Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"product_rule_first_derivative","label":"Product Rule and First Derivative Test","steps":[{"type":"text","order":0,"title":"Apply the product rule","content":"To find the stationary points, we first need to find the first derivative of the function, $\\frac{dy}{dx}$. \n\nThe given curve is:\n$$y = x(x-2)^3$$\n\nWe will use the **product rule**, which states:\n$$\\frac{d}{dx}(uv) = u\\frac{dv}{dx} + v\\frac{du}{dx}$$\n\nLet's define:\n- $u = x \\implies \\frac{du}{dx} = 1$\n- $v = (x-2)^3 \\implies \\frac{dv}{dx} = 3(x-2)^2$ (using the chain rule)\n\nNow, substitute these into the product rule formula:\n$$\\frac{dy}{dx} = x \\cdot 3(x-2)^2 + (x-2)^3 \\cdot 1$$\n$$\\frac{dy}{dx} = 3x(x-2)^2 + (x-2)^3$$"},{"type":"text","order":1,"title":"Factorise the derivative","content":"To easily solve for the stationary points, we factorise $\\frac{dy}{dx}$ by taking out the common factor, $(x-2)^2$:\n\n$$\\begin{aligned} \\frac{dy}{dx} &= (x-2)^2 \\left[ 3x + (x-2) \\right] \\\\ &= (x-2)^2 (4x - 2) \\\\ &= 2(x-2)^2 (2x - 1) \\end{aligned}$$"},{"type":"text","order":2,"title":"Find x-coordinates","content":"Stationary points occur where the gradient is zero, so we set $\\frac{dy}{dx} = 0$:\n\n$$2(x-2)^2 (2x - 1) = 0$$\n\nThis gives us two solutions:\n- From $(x-2)^2 = 0$, we get $x = 2$\n- From $2x - 1 = 0$, we get $x = \\frac{1}{2}$"},{"type":"text","order":3,"title":"Find y-coordinates","content":"Now, we find the corresponding $y$-coordinates by substituting our $x$-values back into the original curve equation, $y = x(x-2)^3$.\n\n**For $x = 2$:**\n$$y = 2(2-2)^3 = 2(0)^3 = 0$$\nThis gives the stationary point: **$(2, 0)$**\n\n**For $x = \\frac{1}{2}$:**\n$$\\begin{aligned} y &= \\frac{1}{2}\\left(\\frac{1}{2}-2\\right)^3 \\\\ &= \\frac{1}{2}\\left(-\\frac{3}{2}\\right)^3 \\\\ &= \\frac{1}{2}\\left(-\\frac{27}{8}\\right) \\\\ &= -\\frac{27}{16} \\end{aligned}$$\nThis gives the stationary point: **$\\left(\\frac{1}{2}, -\\frac{27}{16}\\right)$**"},{"type":"text","order":4,"title":"Perform first derivative test","content":"To determine the nature of each stationary point, we will construct a sign table for the first derivative, $\\frac{dy}{dx} = 2(x-2)^2 (2x - 1)$, by testing $x$-values in the intervals around our stationary points.\n\nOur key $x$-values are $x = \\frac{1}{2}$ and $x = 2$.\n\n| Interval | $x < \\frac{1}{2}$ (e.g., $x=0$) | $x = \\frac{1}{2}$ | $\\frac{1}{2} < x < 2$ (e.g., $x=1$) | $x = 2$ | $x > 2$ (e.g., $x=3$) |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| **Sign of $\\frac{dy}{dx}$** | $2(-2)^2(-1) = -8 \\implies \\boldsymbol{-}$ | $0$ | $2(-1)^2(1) = 2 \\implies \\boldsymbol{+}$ | $0$ | $2(1)^2(5) = 10 \\implies \\boldsymbol{+}$ |\n| **Slope of tangent** | Negative gradient (falling) | Horizontal | Positive gradient (rising) | Horizontal | Positive gradient (rising) |"},{"type":"text","order":5,"title":"Analyse the stationary points","content":"Using the results from our sign table, we can determine the nature of each stationary point:\n\n1. **At the point $\\left(\\frac{1}{2}, -\\frac{27}{16}\\right)$**:\n - Just to the left ($x < \\frac{1}{2}$), the gradient is **negative**.\n - Just to the right ($x > \\frac{1}{2}$), the gradient is **positive**.\n - Since the gradient changes from negative to positive, the curve reaches a bottom point. Therefore, **$\\left(\\frac{1}{2}, -\\frac{27}{16}\\right)$ is a local minimum**.\n\n2. **At the point $(2, 0)$**:\n - Just to the left ($x < 2$), the gradient is **positive**.\n - Just to the right ($x > 2$), the gradient remains **positive**.\n - Since the gradient is positive on both sides of this horizontal tangent, the curve flattens out but continues to rise. Therefore, **$(2, 0)$ is a non-folding point of inflexion**."}]},{"id":"product_rule_second_derivative","label":"Product Rule and Second Derivative Test","steps":[{"type":"text","order":0,"title":"Apply the Product Rule","content":"To find the stationary points, we first need to differentiate the function $y = x(x-2)^3$ with respect to $x$. \n\nWe will use the **Product Rule**, which is given in the formula booklet as:\n$$\\frac{d}{dx}(uv) = u\\frac{dv}{dx} + v\\frac{du}{dx}$$\n\nLet's define:\n* $u = x \\implies \\frac{du}{dx} = 1$\n* $v = (x-2)^3 \\implies \\frac{dv}{dx} = 3(x-2)^2$ (using the Chain Rule)\n\nNow, substituting these into the Product Rule formula:\n$$\\frac{dy}{dx} = x \\cdot 3(x-2)^2 + (x-2)^3 \\cdot 1$$\n$$\\frac{dy}{dx} = 3x(x-2)^2 + (x-2)^3$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y' = \\htmlData{anchor=term1}{\\color{@sky}{(x-2)^2}}\\color{@teal}{[}3x\\color{@teal}{]} + \\htmlData{anchor=term2}{\\color{@sky}{(x-2)^2}}\\color{@teal}{[}(x-2)\\color{@teal}{]}$"},{"latex":"$$y' = \\htmlData{anchor=factored}{\\color{@sky}{(x-2)^2}}\\color{@teal}{[}3x + (x-2)\\color{@teal}{]}$"},{"latex":"$$y' = (x-2)^2(4x-2) = 2(x-2)^2(2x-1)$"}],"highlights":[{"color":"sky","style":"box","anchor":"term1"},{"color":"sky","style":"box","anchor":"term2"},{"color":"sky","style":"box","anchor":"factored"}],"connections":[{"to":"factored","from":"term1","color":"sky","label":"factor out","style":"solid"},{"to":"factored","from":"term2","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Factorise the Derivative","description":"To find the stationary points easily, factor out the common term $(x-2)^2$ from both parts of the derivative."},{"type":"text","order":2,"title":"Find the Stationary Points","content":"Stationary points occur when the first derivative is zero, so we set $\\frac{dy}{dx} = 0$:\n\n$$2(x-2)^2(2x-1) = 0$$\n\nThis gives us two possible $x$-coordinates:\n1. $x - 2 = 0 \\implies x = 2$\n2. $2x - 1 = 0 \\implies x = \\frac{1}{2}$\n\nNow we find the corresponding $y$-coordinates by substituting these $x$-values back into the original curve equation $y = x(x-2)^3$:\n* For $x = 2$:\n $$y = 2(2-2)^3 = 0 \\implies (2, 0)$$\n* For $x = \\frac{1}{2}$:\n $$y = \\frac{1}{2}\\left(\\frac{1}{2}-2\\right)^3 = \\frac{1}{2}\\left(-\\frac{3}{2}\\right)^3 = \\frac{1}{2}\\left(-\\frac{27}{8}\\right) = -\\frac{27}{16} \\implies \\left(\\frac{1}{2}, -\\frac{27}{16}\\right)$$"},{"type":"text","order":3,"title":"Find the Second Derivative","content":"To determine the nature of these stationary points using the **Second Derivative Test**, we need to find $\\frac{d^2y}{dx^2}$. We differentiate $\\frac{dy}{dx} = 2(x-2)^2(2x-1)$ using the product rule again:\n\nLet:\n* $u = 2(x-2)^2 \\implies \\frac{du}{dx} = 4(x-2)$\n* $v = 2x-1 \\implies \\frac{dv}{dx} = 2$\n\nApplying the product rule:\n$$\\frac{d^2y}{dx^2} = 4(x-2)(2x-1) + 2(x-2)^2 \\cdot 2$$\n$$\\frac{d^2y}{dx^2} = 4(x-2)(2x-1) + 4(x-2)^2$$\n\nFactor out $4(x-2)$:\n$$\\frac{d^2y}{dx^2} = 4(x-2)[(2x-1) + (x-2)]$$\n$$\\frac{d^2y}{dx^2} = 4(x-2)(3x-3) = 12(x-2)(x-1)$$"},{"type":"text","order":4,"title":"Test Point (1/2, -27/16)","content":"Substitute $x = \\frac{1}{2}$ into the second derivative formula:\n\n$$\\frac{d^2y}{dx^2} = 12\\left(\\frac{1}{2}-2\\right)\\left(\\frac{1}{2}-1\\right)$$\n$$\\frac{d^2y}{dx^2} = 12\\left(-\\frac{3}{2}\\right)\\left(-\\frac{1}{2}\\right) = 9$$\n\nSince $\\frac{d^2y}{dx^2} = 9 > 0$ (positive), the stationary point $\\left(\\frac{1}{2}, -\\frac{27}{16}\\right)$ is a **local minimum**."},{"type":"text","order":5,"title":"Test Point (2, 0)","content":"Substitute $x = 2$ into the second derivative formula:\n\n$$\\frac{d^2y}{dx^2} = 12(2-2)(2-1) = 0$$\n\nBecause the second derivative is equal to $0$, the second derivative test is inconclusive! We must back it up using a gradient test (first derivative test) for points close to $x = 2$:\n\n* **Just below $x = 2$** (e.g., $x = 1.9$):\n $$\\frac{dy}{dx} = 2(1.9-2)^2(2(1.9)-1) = 2(-0.1)^2(2.8) = 0.056 > 0 \\quad \\text{(positive gradient)}$$\n* **Just above $x = 2$** (e.g., $x = 2.1$):\n $$\\frac{dy}{dx} = 2(2.1-2)^2(2(2.1)-1) = 2(0.1)^2(3.2) = 0.064 > 0 \\quad \\text{(positive gradient)}$$\n\nSince the gradient is positive both before and after the stationary point, $(2, 0)$ is a **point of inflexion**."}]},{"id":"expansion_method","label":"Expansion and Polynomial Differentiation","steps":[{"type":"text","order":0,"title":"Expand the function","content":"To differentiate the function using the **Expansion and Polynomial Differentiation** approach, we first fully expand the expression for $y$.\n\nStart by expanding the cubic term $(x-2)^3$ using the binomial expansion formula or direct expansion:\n$$(x-2)^3 = x^3 - 3(x^2)(2) + 3(x)(2^2) - 2^3 = x^3 - 6x^2 + 12x - 8$$\n\nNow, multiply this by the front factor $x$:\n$$y = x(x^3 - 6x^2 + 12x - 8)$$\n$$y = x^4 - 6x^3 + 12x^2 - 8x$$","highlights":[{"text":"y = x(x-2)^3","location":"specification"}]},{"type":"text","order":1,"title":"Differentiate term-by-term","content":"Now we differentiate the polynomial term-by-term with respect to $x$ using the power rule $\\frac{d}{dx}(x^n) = nx^{n-1}$:\n\n$$\\frac{dy}{dx} = \\frac{d}{dx}(x^4 - 6x^3 + 12x^2 - 8x)$$\n$$\\frac{dy}{dx} = 4x^3 - 18x^2 + 24x - 8$$"},{"type":"text","order":2,"title":"Find x-coordinates","content":"To find the stationary points, we set the derivative equal to zero:\n$$4x^3 - 18x^2 + 24x - 8 = 0$$\n\nWe can factor out a common factor of $2$:\n$$2(2x^3 - 9x^2 + 12x - 4) = 0$$\n\nBy testing small integer values, we can find a root for the cubic polynomial $f(x) = 2x^3 - 9x^2 + 12x - 4$:\n- For $x=2$: $2(2)^3 - 9(2)^2 + 12(2) - 4 = 16 - 36 + 24 - 4 = 0$\n\nSince $x=2$ is a root, $(x-2)$ is a factor. Dividing $2x^3 - 9x^2 + 12x - 4$ by $(x-2)$ via algebraic division or synthetic division yields:\n$$2x^3 - 9x^2 + 12x - 4 = (x-2)(2x^2 - 5x + 2)$$\n\nFactoring the quadratic factor further:\n$$2x^2 - 5x + 2 = (2x-1)(x-2)$$\n\nPutting it all together:\n$$\\frac{dy}{dx} = 2(x-2)^2(2x-1) = 0$$\n\nThus, the $x$-coordinates of our stationary points are:\n$$x = 2 \\quad \\text{and} \\quad x = \\frac{1}{2}$$","calculator":[{"gdc":{"expression":"4x^3 - 18x^2 + 24x - 8 = 0","instructions":"1. Go to the 'Equation' or 'Polynomial' solver menu.\n2. Choose polynomial of degree 3.\n3. Enter the coefficients: 4, -18, 24, -8.\n4. Solve to get the roots."},"ti84":{"expression":"4x^3 - 18x^2 + 24x - 8 = 0","instructions":"1. Press [APPS] and select 'PlySmlt2'.\n2. Select 'Polynomial Root Finder'.\n3. Set Order to 3 and press [NEXT] (GRAPH).\n4. Enter the coefficients: a3 = 4, a2 = -18, a1 = 24, a0 = -8.\n5. Press [SOLVE] (GRAPH)."},"desmos":{"expression":"4x^3 - 18x^2 + 24x - 8 = 0","instructions":"Type `y = 4x^3 - 18x^2 + 24x - 8` and identify where the graph crosses the x-axis."},"description":"You can find the roots of the cubic equation $4x^3 - 18x^2 + 24x - 8 = 0$ using your graphing calculator's polynomial solver."}]},{"type":"text","order":3,"title":"Calculate y-coordinates","content":"Substitute the $x$-values back into the original curve equation $y = x(x-2)^3$ to find the corresponding $y$-coordinates.\n\n- **For $x=2$:**\n $$y = 2(2-2)^3 = 2(0)^3 = 0$$\n This gives the stationary point: $$(2, 0)$$\n\n- **For $x=\\frac{1}{2}$:**\n $$y = \\frac{1}{2}\\left(\\frac{1}{2}-2\\right)^3 = \\frac{1}{2}\\left(-\\frac{3}{2}\\right)^3 = \\frac{1}{2}\\left(-\\frac{27}{8}\\right) = -\\frac{27}{16}$$\n This gives the stationary point: $$\\left(\\frac{1}{2}, -\\frac{27}{16}\\right)$$\n\nThese coordinate pairs match the marking criteria directly."},{"type":"text","order":4,"title":"Determine the second derivative","content":"To determine the nature of these stationary points, we can compute the second derivative $\\frac{d^2y}{dx^2}$ by differentiating our polynomial expression for $\\frac{dy}{dx} = 4x^3 - 18x^2 + 24x - 8$:\n\n$$\\frac{d^2y}{dx^2} = \\frac{d}{dx}(4x^3 - 18x^2 + 24x - 8)$$\n$$\\frac{d^2y}{dx^2} = 12x^2 - 36x + 24$$"},{"type":"text","order":5,"title":"Analyze the stationary points","content":"Let's test each stationary point using the second derivative:\n\n1. **At $x = \\frac{1}{2}$:**\n $$\\frac{d^2y}{dx^2} = 12\\left(\\frac{1}{2}\\right)^2 - 36\\left(\\frac{1}{2}\\right) + 24 = 12\\left(\\frac{1}{4}\\right) - 18 + 24 = 3 - 18 + 24 = 9$$\n Since $\\frac{d^2y}{dx^2} = 9 > 0$, the stationary point $\\left(\\frac{1}{2}, -\\frac{27}{16}\\right)$ is a **Local Minimum**.\n\n2. **At $x = 2$:**\n $$\\frac{d^2y}{dx^2} = 12(2)^2 - 36(2) + 24 = 48 - 72 + 24 = 0$$\n Because the second derivative is zero, the second derivative test is inconclusive. We must look at the sign of the first derivative $\\frac{dy}{dx} = 2(x-2)^2(2x-1)$ just to the left and right of $x = 2$:\n - Choose $x = 1.9$: $\\frac{dy}{dx} = 2(1.9-2)^2(2(1.9)-1) = 2(-0.1)^2(2.8) > 0$ (positive)\n - Choose $x = 2.1$: $\\frac{dy}{dx} = 2(2.1-2)^2(2(2.1)-1) = 2(0.1)^2(3.2) > 0$ (positive)\n\n Since the gradient does not change sign as it passes through $x = 2$ (it remains positive on both sides), the point $(2, 0)$ is a **Point of Inflexion**."}]}],"generatedAt":"2026-06-09T01:59:55.825Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"4209ae28-8832-4e0e-bf3c-2db51bcffb57:product_rule_first_derivative","label":"Product Rule and First Derivative Test","steps":[{"type":"text","order":0,"title":"Apply the product rule","content":"To find the stationary points, we first need to find the first derivative of the function, $\\frac{dy}{dx}$. \n\nThe given curve is:\n$$y = x(x-2)^3$$\n\nWe will use the **product rule**, which states:\n$$\\frac{d}{dx}(uv) = u\\frac{dv}{dx} + v\\frac{du}{dx}$$\n\nLet's define:\n- $u = x \\implies \\frac{du}{dx} = 1$\n- $v = (x-2)^3 \\implies \\frac{dv}{dx} = 3(x-2)^2$ (using the chain rule)\n\nNow, substitute these into the product rule formula:\n$$\\frac{dy}{dx} = x \\cdot 3(x-2)^2 + (x-2)^3 \\cdot 1$$\n$$\\frac{dy}{dx} = 3x(x-2)^2 + (x-2)^3$$"},{"type":"text","order":1,"title":"Factorise the derivative","content":"To easily solve for the stationary points, we factorise $\\frac{dy}{dx}$ by taking out the common factor, $(x-2)^2$:\n\n$$\\begin{aligned} \\frac{dy}{dx} &= (x-2)^2 \\left[ 3x + (x-2) \\right] \\\\ &= (x-2)^2 (4x - 2) \\\\ &= 2(x-2)^2 (2x - 1) \\end{aligned}$$"},{"type":"text","order":2,"title":"Find x-coordinates","content":"Stationary points occur where the gradient is zero, so we set $\\frac{dy}{dx} = 0$:\n\n$$2(x-2)^2 (2x - 1) = 0$$\n\nThis gives us two solutions:\n- From $(x-2)^2 = 0$, we get $x = 2$\n- From $2x - 1 = 0$, we get $x = \\frac{1}{2}$"},{"type":"text","order":3,"title":"Find y-coordinates","content":"Now, we find the corresponding $y$-coordinates by substituting our $x$-values back into the original curve equation, $y = x(x-2)^3$.\n\n**For $x = 2$:**\n$$y = 2(2-2)^3 = 2(0)^3 = 0$$\nThis gives the stationary point: **$(2, 0)$**\n\n**For $x = \\frac{1}{2}$:**\n$$\\begin{aligned} y &= \\frac{1}{2}\\left(\\frac{1}{2}-2\\right)^3 \\\\ &= \\frac{1}{2}\\left(-\\frac{3}{2}\\right)^3 \\\\ &= \\frac{1}{2}\\left(-\\frac{27}{8}\\right) \\\\ &= -\\frac{27}{16} \\end{aligned}$$\nThis gives the stationary point: **$\\left(\\frac{1}{2}, -\\frac{27}{16}\\right)$**"},{"type":"text","order":4,"title":"Perform first derivative test","content":"To determine the nature of each stationary point, we will construct a sign table for the first derivative, $\\frac{dy}{dx} = 2(x-2)^2 (2x - 1)$, by testing $x$-values in the intervals around our stationary points.\n\nOur key $x$-values are $x = \\frac{1}{2}$ and $x = 2$.\n\n| Interval | $x < \\frac{1}{2}$ (e.g., $x=0$) | $x = \\frac{1}{2}$ | $\\frac{1}{2} < x < 2$ (e.g., $x=1$) | $x = 2$ | $x > 2$ (e.g., $x=3$) |\n| :--- | :--- | :--- | :--- | :--- | :--- |\n| **Sign of $\\frac{dy}{dx}$** | $2(-2)^2(-1) = -8 \\implies \\boldsymbol{-}$ | $0$ | $2(-1)^2(1) = 2 \\implies \\boldsymbol{+}$ | $0$ | $2(1)^2(5) = 10 \\implies \\boldsymbol{+}$ |\n| **Slope of tangent** | Negative gradient (falling) | Horizontal | Positive gradient (rising) | Horizontal | Positive gradient (rising) |"},{"type":"text","order":5,"title":"Analyse the stationary points","content":"Using the results from our sign table, we can determine the nature of each stationary point:\n\n1. **At the point $\\left(\\frac{1}{2}, -\\frac{27}{16}\\right)$**:\n - Just to the left ($x < \\frac{1}{2}$), the gradient is **negative**.\n - Just to the right ($x > \\frac{1}{2}$), the gradient is **positive**.\n - Since the gradient changes from negative to positive, the curve reaches a bottom point. Therefore, **$\\left(\\frac{1}{2}, -\\frac{27}{16}\\right)$ is a local minimum**.\n\n2. **At the point $(2, 0)$**:\n - Just to the left ($x < 2$), the gradient is **positive**.\n - Just to the right ($x > 2$), the gradient remains **positive**.\n - Since the gradient is positive on both sides of this horizontal tangent, the curve flattens out but continues to rise. Therefore, **$(2, 0)$ is a non-folding point of inflexion**."}]},{"id":"4209ae28-8832-4e0e-bf3c-2db51bcffb57:product_rule_second_derivative","label":"Product Rule and Second Derivative Test","steps":[{"type":"text","order":0,"title":"Apply the Product Rule","content":"To find the stationary points, we first need to differentiate the function $y = x(x-2)^3$ with respect to $x$. \n\nWe will use the **Product Rule**, which is given in the formula booklet as:\n$$\\frac{d}{dx}(uv) = u\\frac{dv}{dx} + v\\frac{du}{dx}$$\n\nLet's define:\n* $u = x \\implies \\frac{du}{dx} = 1$\n* $v = (x-2)^3 \\implies \\frac{dv}{dx} = 3(x-2)^2$ (using the Chain Rule)\n\nNow, substituting these into the Product Rule formula:\n$$\\frac{dy}{dx} = x \\cdot 3(x-2)^2 + (x-2)^3 \\cdot 1$$\n$$\\frac{dy}{dx} = 3x(x-2)^2 + (x-2)^3$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$y' = \\htmlData{anchor=term1}{\\color{@sky}{(x-2)^2}}\\color{@teal}{[}3x\\color{@teal}{]} + \\htmlData{anchor=term2}{\\color{@sky}{(x-2)^2}}\\color{@teal}{[}(x-2)\\color{@teal}{]}$"},{"latex":"$$y' = \\htmlData{anchor=factored}{\\color{@sky}{(x-2)^2}}\\color{@teal}{[}3x + (x-2)\\color{@teal}{]}$"},{"latex":"$$y' = (x-2)^2(4x-2) = 2(x-2)^2(2x-1)$"}],"highlights":[{"color":"sky","style":"box","anchor":"term1"},{"color":"sky","style":"box","anchor":"term2"},{"color":"sky","style":"box","anchor":"factored"}],"connections":[{"to":"factored","from":"term1","color":"sky","label":"factor out","style":"solid"},{"to":"factored","from":"term2","color":"sky","label":null,"style":"solid"}]},"order":1,"title":"Factorise the Derivative","description":"To find the stationary points easily, factor out the common term $(x-2)^2$ from both parts of the derivative."},{"type":"text","order":2,"title":"Find the Stationary Points","content":"Stationary points occur when the first derivative is zero, so we set $\\frac{dy}{dx} = 0$:\n\n$$2(x-2)^2(2x-1) = 0$$\n\nThis gives us two possible $x$-coordinates:\n1. $x - 2 = 0 \\implies x = 2$\n2. $2x - 1 = 0 \\implies x = \\frac{1}{2}$\n\nNow we find the corresponding $y$-coordinates by substituting these $x$-values back into the original curve equation $y = x(x-2)^3$:\n* For $x = 2$:\n $$y = 2(2-2)^3 = 0 \\implies (2, 0)$$\n* For $x = \\frac{1}{2}$:\n $$y = \\frac{1}{2}\\left(\\frac{1}{2}-2\\right)^3 = \\frac{1}{2}\\left(-\\frac{3}{2}\\right)^3 = \\frac{1}{2}\\left(-\\frac{27}{8}\\right) = -\\frac{27}{16} \\implies \\left(\\frac{1}{2}, -\\frac{27}{16}\\right)$$"},{"type":"text","order":3,"title":"Find the Second Derivative","content":"To determine the nature of these stationary points using the **Second Derivative Test**, we need to find $\\frac{d^2y}{dx^2}$. We differentiate $\\frac{dy}{dx} = 2(x-2)^2(2x-1)$ using the product rule again:\n\nLet:\n* $u = 2(x-2)^2 \\implies \\frac{du}{dx} = 4(x-2)$\n* $v = 2x-1 \\implies \\frac{dv}{dx} = 2$\n\nApplying the product rule:\n$$\\frac{d^2y}{dx^2} = 4(x-2)(2x-1) + 2(x-2)^2 \\cdot 2$$\n$$\\frac{d^2y}{dx^2} = 4(x-2)(2x-1) + 4(x-2)^2$$\n\nFactor out $4(x-2)$:\n$$\\frac{d^2y}{dx^2} = 4(x-2)[(2x-1) + (x-2)]$$\n$$\\frac{d^2y}{dx^2} = 4(x-2)(3x-3) = 12(x-2)(x-1)$$"},{"type":"text","order":4,"title":"Test Point (1/2, -27/16)","content":"Substitute $x = \\frac{1}{2}$ into the second derivative formula:\n\n$$\\frac{d^2y}{dx^2} = 12\\left(\\frac{1}{2}-2\\right)\\left(\\frac{1}{2}-1\\right)$$\n$$\\frac{d^2y}{dx^2} = 12\\left(-\\frac{3}{2}\\right)\\left(-\\frac{1}{2}\\right) = 9$$\n\nSince $\\frac{d^2y}{dx^2} = 9 > 0$ (positive), the stationary point $\\left(\\frac{1}{2}, -\\frac{27}{16}\\right)$ is a **local minimum**."},{"type":"text","order":5,"title":"Test Point (2, 0)","content":"Substitute $x = 2$ into the second derivative formula:\n\n$$\\frac{d^2y}{dx^2} = 12(2-2)(2-1) = 0$$\n\nBecause the second derivative is equal to $0$, the second derivative test is inconclusive! We must back it up using a gradient test (first derivative test) for points close to $x = 2$:\n\n* **Just below $x = 2$** (e.g., $x = 1.9$):\n $$\\frac{dy}{dx} = 2(1.9-2)^2(2(1.9)-1) = 2(-0.1)^2(2.8) = 0.056 > 0 \\quad \\text{(positive gradient)}$$\n* **Just above $x = 2$** (e.g., $x = 2.1$):\n $$\\frac{dy}{dx} = 2(2.1-2)^2(2(2.1)-1) = 2(0.1)^2(3.2) = 0.064 > 0 \\quad \\text{(positive gradient)}$$\n\nSince the gradient is positive both before and after the stationary point, $(2, 0)$ is a **point of inflexion**."}]},{"id":"4209ae28-8832-4e0e-bf3c-2db51bcffb57:expansion_method","label":"Expansion and Polynomial Differentiation","steps":[{"type":"text","order":0,"title":"Expand the function","content":"To differentiate the function using the **Expansion and Polynomial Differentiation** approach, we first fully expand the expression for $y$.\n\nStart by expanding the cubic term $(x-2)^3$ using the binomial expansion formula or direct expansion:\n$$(x-2)^3 = x^3 - 3(x^2)(2) + 3(x)(2^2) - 2^3 = x^3 - 6x^2 + 12x - 8$$\n\nNow, multiply this by the front factor $x$:\n$$y = x(x^3 - 6x^2 + 12x - 8)$$\n$$y = x^4 - 6x^3 + 12x^2 - 8x$$","highlights":[{"text":"y = x(x-2)^3","location":"specification"}]},{"type":"text","order":1,"title":"Differentiate term-by-term","content":"Now we differentiate the polynomial term-by-term with respect to $x$ using the power rule $\\frac{d}{dx}(x^n) = nx^{n-1}$:\n\n$$\\frac{dy}{dx} = \\frac{d}{dx}(x^4 - 6x^3 + 12x^2 - 8x)$$\n$$\\frac{dy}{dx} = 4x^3 - 18x^2 + 24x - 8$$"},{"type":"text","order":2,"title":"Find x-coordinates","content":"To find the stationary points, we set the derivative equal to zero:\n$$4x^3 - 18x^2 + 24x - 8 = 0$$\n\nWe can factor out a common factor of $2$:\n$$2(2x^3 - 9x^2 + 12x - 4) = 0$$\n\nBy testing small integer values, we can find a root for the cubic polynomial $f(x) = 2x^3 - 9x^2 + 12x - 4$:\n- For $x=2$: $2(2)^3 - 9(2)^2 + 12(2) - 4 = 16 - 36 + 24 - 4 = 0$\n\nSince $x=2$ is a root, $(x-2)$ is a factor. Dividing $2x^3 - 9x^2 + 12x - 4$ by $(x-2)$ via algebraic division or synthetic division yields:\n$$2x^3 - 9x^2 + 12x - 4 = (x-2)(2x^2 - 5x + 2)$$\n\nFactoring the quadratic factor further:\n$$2x^2 - 5x + 2 = (2x-1)(x-2)$$\n\nPutting it all together:\n$$\\frac{dy}{dx} = 2(x-2)^2(2x-1) = 0$$\n\nThus, the $x$-coordinates of our stationary points are:\n$$x = 2 \\quad \\text{and} \\quad x = \\frac{1}{2}$$","calculator":[{"gdc":{"expression":"4x^3 - 18x^2 + 24x - 8 = 0","instructions":"1. Go to the 'Equation' or 'Polynomial' solver menu.\n2. Choose polynomial of degree 3.\n3. Enter the coefficients: 4, -18, 24, -8.\n4. Solve to get the roots."},"ti84":{"expression":"4x^3 - 18x^2 + 24x - 8 = 0","instructions":"1. Press [APPS] and select 'PlySmlt2'.\n2. Select 'Polynomial Root Finder'.\n3. Set Order to 3 and press [NEXT] (GRAPH).\n4. Enter the coefficients: a3 = 4, a2 = -18, a1 = 24, a0 = -8.\n5. Press [SOLVE] (GRAPH)."},"desmos":{"expression":"4x^3 - 18x^2 + 24x - 8 = 0","instructions":"Type `y = 4x^3 - 18x^2 + 24x - 8` and identify where the graph crosses the x-axis."},"description":"You can find the roots of the cubic equation $4x^3 - 18x^2 + 24x - 8 = 0$ using your graphing calculator's polynomial solver."}]},{"type":"text","order":3,"title":"Calculate y-coordinates","content":"Substitute the $x$-values back into the original curve equation $y = x(x-2)^3$ to find the corresponding $y$-coordinates.\n\n- **For $x=2$:**\n $$y = 2(2-2)^3 = 2(0)^3 = 0$$\n This gives the stationary point: $$(2, 0)$$\n\n- **For $x=\\frac{1}{2}$:**\n $$y = \\frac{1}{2}\\left(\\frac{1}{2}-2\\right)^3 = \\frac{1}{2}\\left(-\\frac{3}{2}\\right)^3 = \\frac{1}{2}\\left(-\\frac{27}{8}\\right) = -\\frac{27}{16}$$\n This gives the stationary point: $$\\left(\\frac{1}{2}, -\\frac{27}{16}\\right)$$\n\nThese coordinate pairs match the marking criteria directly."},{"type":"text","order":4,"title":"Determine the second derivative","content":"To determine the nature of these stationary points, we can compute the second derivative $\\frac{d^2y}{dx^2}$ by differentiating our polynomial expression for $\\frac{dy}{dx} = 4x^3 - 18x^2 + 24x - 8$:\n\n$$\\frac{d^2y}{dx^2} = \\frac{d}{dx}(4x^3 - 18x^2 + 24x - 8)$$\n$$\\frac{d^2y}{dx^2} = 12x^2 - 36x + 24$$"},{"type":"text","order":5,"title":"Analyze the stationary points","content":"Let's test each stationary point using the second derivative:\n\n1. **At $x = \\frac{1}{2}$:**\n $$\\frac{d^2y}{dx^2} = 12\\left(\\frac{1}{2}\\right)^2 - 36\\left(\\frac{1}{2}\\right) + 24 = 12\\left(\\frac{1}{4}\\right) - 18 + 24 = 3 - 18 + 24 = 9$$\n Since $\\frac{d^2y}{dx^2} = 9 > 0$, the stationary point $\\left(\\frac{1}{2}, -\\frac{27}{16}\\right)$ is a **Local Minimum**.\n\n2. **At $x = 2$:**\n $$\\frac{d^2y}{dx^2} = 12(2)^2 - 36(2) + 24 = 48 - 72 + 24 = 0$$\n Because the second derivative is zero, the second derivative test is inconclusive. We must look at the sign of the first derivative $\\frac{dy}{dx} = 2(x-2)^2(2x-1)$ just to the left and right of $x = 2$:\n - Choose $x = 1.9$: $\\frac{dy}{dx} = 2(1.9-2)^2(2(1.9)-1) = 2(-0.1)^2(2.8) > 0$ (positive)\n - Choose $x = 2.1$: $\\frac{dy}{dx} = 2(2.1-2)^2(2(2.1)-1) = 2(0.1)^2(3.2) > 0$ (positive)\n\n Since the gradient does not change sign as it passes through $x = 2$ (it remains positive on both sides), the point $(2, 0)$ is a **Point of Inflexion**."}]}],"generatedAt":"2026-06-09T02:00:17.928Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"36a3d7b5-ff9e-4cb5-8a36-77923b67b622","revisionId":"15c2fd77-533f-4cb4-b4c3-7ac96a15e486","questionNumber":"5","questionType":"LA","specification":"","difficulty":3,"difficultyBand":"hard","tier":null,"totalMarks":8,"sourceTitle":"9.4 The Product Rule","calculatorUse":null,"isCalculation":true,"parts":[{"id":"ed70ef43-3f43-4739-a177-9f1ea527098b","label":"0","content":"Find the coordinates of the stationary points of the curve $y = x(x-3)^3$ and determine the nature of the stationary points.","markscheme":"- Differentiates using product rule: $\\frac{dy}{dx} = (x-3)^3 + 3x(x-3)^2$ M1\n- Factors to $(x-3)^2(4x-3)$ M1\n- Sets $\\frac{dy}{dx} = 0$ to find $x = 3$ and $x = \\frac{3}{4}$ M1\n- Finds $y$-coordinates: $(3, 0)$ and $(\\frac{3}{4}, -\\frac{2187}{256})$ A1 A1\n- Uses second derivative or gradient testing to determine nature M1\n- Identifies $(\\frac{3}{4}, -\\frac{2187}{256})$ as a Minimum A1\n- Identifies $(3, 0)$ as a Point of Inflexion A1","marks":8,"order":0,"markingRubric":{"totalMarks":8,"points":[{"id":"p1","marks":1,"description":"Correctly differentiates using the product rule.","code":"M"},{"id":"p2","marks":1,"description":"Factors the derivative to find stationary points.","code":"M"},{"id":"p3","marks":1,"description":"Identifies $x$-coordinates $3$ and $0.75$.","code":"M"},{"id":"p4","marks":1,"description":"Correct $y$-coordinate for $x=3$ is $0$.","code":"A"},{"id":"p5","marks":1,"description":"Correct $y$-coordinate for $x=0.75$ is $-2187/256$.","code":"A"},{"id":"p6","marks":1,"description":"Method to determine nature (e.g., second derivative or checking gradients).","code":"M"},{"id":"p7","marks":1,"description":"Correctly identifies the minimum point.","code":"A"},{"id":"p8","marks":1,"description":"Correctly identifies the point of inflexion.","code":"A"}],"finalAnswers":["(3, 0)","(\\frac{3}{4}, -\\frac{2187}{256})"],"generatedAt":"2026-05-06T03:36:51.420Z","modelVersion":"gemini-3-flash-preview"},"tableImages":[],"explanation":{"methods":[{"id":"product_rule_first_derivative","label":"Product Rule and First Derivative Test","steps":[{"type":"text","order":0,"title":"Identify the product components","content":"To find the stationary points of the curve $y = x(x-3)^3$, we first need to find its derivative $\\frac{dy}{dx}$. \n\nSince the equation is a product of two terms, we can use the **Product Rule**:\n\n$$\\frac{d}{dx}(uv) = u\\frac{dv}{dx} + v\\frac{du}{dx}$$\n\nLet's define our parts:\n* $u = x \\implies \\frac{du}{dx} = 1$\n* $v = (x-3)^3$\n\nTo find $\\frac{dv}{dx}$, we use the **Chain Rule**:\n$$\\frac{dv}{dx} = 3(x-3)^2 \\cdot 1 = 3(x-3)^2$$"},{"type":"text","order":1,"title":"Apply the Product Rule","content":"Now, we substitute our parts into the Product Rule formula:\n\n$$\\frac{dy}{dx} = x \\cdot [3(x-3)^2] + (x-3)^3 \\cdot [1]$$\n\nSimplifying this gives:\n\n$$\\frac{dy}{dx} = 3x(x-3)^2 + (x-3)^3$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{dy}{dx} = 3x\\htmlData{anchor=term1}{\\color{@sky}{(x-3)^2}} + \\htmlData{anchor=term2}{\\color{@sky}{(x-3)^3}}$"},{"latex":"$$\\phantom{\\frac{dy}{dx}} = \\htmlData{anchor=factor}{\\color{@sky}{(x-3)^2}} \\left[ 3x + \\htmlData{anchor=rem}{(x-3)} \\right]$"},{"latex":"$$\\phantom{\\frac{dy}{dx}} = (x-3)^2(4x - 3)$"}],"highlights":[{"color":"sky","style":"underline","anchor":"term1"},{"color":"sky","style":"underline","anchor":"term2"},{"color":"sky","style":"box","anchor":"factor"}],"connections":[{"to":"factor","from":"term1","color":"sky","label":"factor out","style":"solid"},{"to":"rem","from":"term2","color":"sky","label":"remaining power","style":"solid"}]},"order":2,"title":"Factorise the derivative","description":"To easily find the stationary points, we factorise the derivative by pulling out the common term $(x-3)^2$."},{"type":"text","order":3,"title":"Find the stationary x-values","content":"Stationary points occur when the gradient of the curve is zero:\n\n$$\\frac{dy}{dx} = 0$$\n\nUsing our factorised derivative:\n\n$$(x-3)^2(4x-3) = 0$$\n\nSetting each factor to zero gives:\n\n$$\\begin{aligned} (x-3)^2 = 0 &\\implies x = 3 \\\\ 4x-3 = 0 &\\implies x = \\frac{3}{4} \\end{aligned}$$"},{"type":"text","order":4,"title":"Calculate the y-coordinates","content":"Now we substitute our $x$-values back into the original curve equation $y = x(x-3)^3$ to find the corresponding $y$-coordinates.\n\n**For $x = 3$:**\n$$y = 3(3-3)^3 = 3(0)^3 = 0$$\nThis gives the stationary point **$(3, 0)$**.\n\n**For $x = \\frac{3}{4}$:**\n$$\\begin{aligned} y &= \\frac{3}{4}\\left(\\frac{3}{4}-3\\right)^3 \\\\ y &= \\frac{3}{4}\\left(-\\frac{9}{4}\\right)^3 \\\\ y &= \\frac{3}{4}\\left(-\\frac{729}{64}\\right) \\\\ y &= -\\frac{2187}{256} \\end{aligned}$$\nThis gives the stationary point **$\\left(\\frac{3}{4}, -\\frac{2187}{256}\\right)$**."},{"type":"text","order":5,"title":"Determine nature using sign table","content":"We will use the **First Derivative Test** by constructing a gradient sign table. We test the sign of $\\frac{dy}{dx} = (x-3)^2(4x-3)$ around our stationary points $x = 0.75$ and $x = 3$.\n\nLet's choose convenient test values:\n\n| $x$ | $x < 0.75$ (e.g., $0$) | $x = 0.75$ | $0.75 < x < 3$ (e.g., $1$) | $x = 3$ | $x > 3$ (e.g., $4$) |\n| :--- | :---: | :---: | :---: | :---: | :---: |\n| **Sign of $(x-3)^2$** | $+$ | $0$ | $+$ | $0$ | $+$ |\n| **Sign of $(4x-3)$** | $-$ | $0$ | $+$ | $+$ | $+$ |\n| **Sign of $\\frac{dy}{dx}$** | $-$ | $0$ | $+$ | $0$ | $+$ |\n| **Shape of curve** | $\\backslash$ | $\\_$ | $/$ | $\\_$ | $/$ |\n\nFrom the gradient patterns:\n* Around $x = \\frac{3}{4}$, the gradient changes from negative (decreasing) to positive (increasing), indicating a **local minimum**.\n* Around $x = 3$, the gradient stays positive (increasing) on both sides, indicating a **point of inflexion**."},{"type":"text","order":6,"title":"State the final conclusion","content":"We have successfully found and classified both stationary points:\n\n* **$\\left(\\frac{3}{4}, -\\frac{2187}{256}\\right)$** is a **local minimum**.\n* **$(3, 0)$** is a **point of inflexion**.\n\n### How marks are awarded:\n* **M1** for applying the Product Rule to get $\\frac{dy}{dx} = (x-3)^3 + 3x(x-3)^2$.\n* **M1** for factorising it to $(x-3)^2(4x-3)$.\n* **M1** for setting the derivative to $0$ and finding $x = 3$ and $x = \\frac{3}{4}$.\n* **A2** (1 mark each) for correctly calculating the $y$-coordinates: $(3,0)$ and $\\left(\\frac{3}{4}, -\\frac{2187}{256}\\right)$.\n* **M1** for setting up a gradient sign test around both stationary points.\n* **A2** (1 mark each) for correctly identifying the minimum and the point of inflexion."}]},{"id":"expansion_second_derivative","label":"Expansion and Second Derivative Test","steps":[{"type":"text","order":0,"title":"Expand the function","content":"To follow our chosen approach, we will first expand the expression for $y$ into a single polynomial. Let's start by expanding $(x-3)^3$ using the binomial expansion:\n\n$$(x-3)^3 = x^3 - 9x^2 + 27x - 27$$\n\nNow, multiply this result by $x$ to get the fully expanded polynomial form of the curve:\n\n$$y = x(x^3 - 9x^2 + 27x - 27) = x^4 - 9x^3 + 27x^2 - 27x$$"},{"type":"text","order":1,"title":"Find the first derivative","content":"Now we differentiate $y$ term-by-term with respect to $x$:\n\n$$\\frac{dy}{dx} = 4x^3 - 27x^2 + 54x - 27$$"},{"type":"text","order":2,"title":"Find the stationary points","content":"Stationary points occur when the gradient is zero, so we set $\\frac{dy}{dx} = 0$:\n\n$$4x^3 - 27x^2 + 54x - 27 = 0$$\n\nWe can factorise this polynomial. Since we know $(x-3)$ is likely a factor from the original expression, we can check if $x=3$ is a root: \n$$4(3)^3 - 27(3)^2 + 54(3) - 27 = 108 - 243 + 162 - 27 = 0$$\n\nThus, $(x-3)$ is a factor. Dividing the cubic by $(x-3)$, or looking for a common factor, we can rewrite it as:\n\n$$(x-3)^2(4x-3) = 0$$\n\nThis gives two solutions for $x$:\n\n$$x = 3 \\quad \\text{and} \\quad x = \\frac{3}{4}$$"},{"type":"text","order":3,"title":"Calculate the y-coordinates","content":"Substitute each $x$-value back into the original curve equation $y = x(x-3)^3$:\n\n- For $x = 3$:\n $$y = 3(3-3)^3 = 0$$\n This gives the stationary point $\\mathbf{(3, 0)}$.\n\n- For $x = \\frac{3}{4}$:\n $$y = \\frac{3}{4}\\left(\\frac{3}{4}-3\\right)^3 = \\frac{3}{4}\\left(-\\frac{9}{4}\\right)^3 = \\frac{3}{4}\\left(-\\frac{729}{64}\\right) = -\\frac{2187}{256}$$\n This gives the stationary point $\\mathbf{\\left(\\frac{3}{4}, -\\frac{2187}{256}\\right)}$."},{"type":"text","order":4,"title":"Find the second derivative","content":"To determine the nature of these points, we find the second derivative, $\\frac{d^2y}{dx^2}$, by differentiating our first derivative expression:\n\n$$\\frac{dy}{dx} = 4x^3 - 27x^2 + 54x - 27$$\n\n$$\\frac{d^2y}{dx^2} = 12x^2 - 54x + 54$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Substitute } \\htmlData{anchor=xval1}{\\color{@sky}{x = \\frac{3}{4}}} \\text{ into } \\frac{d^2y}{dx^2} = 12\\htmlData{anchor=xsub1}{\\color{@sky}{x}}^2 - 54\\htmlData{anchor=xsub2}{\\color{@sky}{x}} + 54:$"},{"latex":"$$$\\begin{aligned} \\frac{d^2y}{dx^2} &= 12\\left(\\color{@sky}{\\frac{3}{4}}\\right)^2 - 54\\left(\\color{@sky}{\\frac{3}{4}}\\right) + 54 \\\\ &= 12\\left(\\frac{9}{16}\\right) - \\frac{81}{2} + 54 \\\\ &= \\frac{27}{4} - \\frac{162}{4} + \\frac{216}{4} \\\\ &= \\htmlData{anchor=d2yval}{\\color{@green}{\\frac{81}{4}}} > 0 \\end{aligned}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"xval1"},{"color":"green","style":"box","anchor":"d2yval"}],"connections":[{"to":"xsub1","from":"xval1","color":"sky","label":"substitute","style":"solid"}]},"order":5,"title":"Test the first point","description":"Since the second derivative is positive at $x = \\frac{3}{4}$, this point is a local minimum."},{"type":"text","order":6,"title":"Test the second point","content":"Substitute $x = 3$ into the second derivative:\n\n$$\\frac{d^2y}{dx^2} = 12(3)^2 - 54(3) + 54 = 108 - 162 + 54 = 0$$\n\nSince $\\frac{d^2y}{dx^2} = 0$, the second derivative test is inconclusive on its own. We must check for a change in the sign of concavity (the sign of $\\frac{d^2y}{dx^2}$) just to the left and right of $x = 3$:\n\n- Let's test a point slightly to the left, $x = 2.9$:\n $$\\frac{d^2y}{dx^2} = 12(2.9)^2 - 54(2.9) + 54 = 100.92 - 156.6 + 54 = -1.68 < 0 \\quad \\text{(concave down)}$$\n- Let's test a point slightly to the right, $x = 3.1$:\n $$\\frac{d^2y}{dx^2} = 12(3.1)^2 - 54(3.1) + 54 = 115.32 - 167.4 + 54 = 1.92 > 0 \\quad \\text{(concave up)}$$\n\nBecause the second derivative changes sign from negative to positive through $x = 3$, the point $(3, 0)$ is a **point of inflexion**."},{"type":"text","order":7,"title":"Summarise the results","content":"Our final classifications for the stationary points are:\n- $\\mathbf{\\left(\\frac{3}{4}, -\\frac{2187}{256}\\right)}$ is a **local minimum**.\n- $\\mathbf{(3, 0)}$ is a **point of inflexion**."}]}],"generatedAt":"2026-06-09T05:04:11.694Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"options":[]}],"options":[],"explanation":{"methods":[{"id":"ed70ef43-3f43-4739-a177-9f1ea527098b:product_rule_first_derivative","label":"Product Rule and First Derivative Test","steps":[{"type":"text","order":0,"title":"Identify the product components","content":"To find the stationary points of the curve $y = x(x-3)^3$, we first need to find its derivative $\\frac{dy}{dx}$. \n\nSince the equation is a product of two terms, we can use the **Product Rule**:\n\n$$\\frac{d}{dx}(uv) = u\\frac{dv}{dx} + v\\frac{du}{dx}$$\n\nLet's define our parts:\n* $u = x \\implies \\frac{du}{dx} = 1$\n* $v = (x-3)^3$\n\nTo find $\\frac{dv}{dx}$, we use the **Chain Rule**:\n$$\\frac{dv}{dx} = 3(x-3)^2 \\cdot 1 = 3(x-3)^2$$"},{"type":"text","order":1,"title":"Apply the Product Rule","content":"Now, we substitute our parts into the Product Rule formula:\n\n$$\\frac{dy}{dx} = x \\cdot [3(x-3)^2] + (x-3)^3 \\cdot [1]$$\n\nSimplifying this gives:\n\n$$\\frac{dy}{dx} = 3x(x-3)^2 + (x-3)^3$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\frac{dy}{dx} = 3x\\htmlData{anchor=term1}{\\color{@sky}{(x-3)^2}} + \\htmlData{anchor=term2}{\\color{@sky}{(x-3)^3}}$"},{"latex":"$$\\phantom{\\frac{dy}{dx}} = \\htmlData{anchor=factor}{\\color{@sky}{(x-3)^2}} \\left[ 3x + \\htmlData{anchor=rem}{(x-3)} \\right]$"},{"latex":"$$\\phantom{\\frac{dy}{dx}} = (x-3)^2(4x - 3)$"}],"highlights":[{"color":"sky","style":"underline","anchor":"term1"},{"color":"sky","style":"underline","anchor":"term2"},{"color":"sky","style":"box","anchor":"factor"}],"connections":[{"to":"factor","from":"term1","color":"sky","label":"factor out","style":"solid"},{"to":"rem","from":"term2","color":"sky","label":"remaining power","style":"solid"}]},"order":2,"title":"Factorise the derivative","description":"To easily find the stationary points, we factorise the derivative by pulling out the common term $(x-3)^2$."},{"type":"text","order":3,"title":"Find the stationary x-values","content":"Stationary points occur when the gradient of the curve is zero:\n\n$$\\frac{dy}{dx} = 0$$\n\nUsing our factorised derivative:\n\n$$(x-3)^2(4x-3) = 0$$\n\nSetting each factor to zero gives:\n\n$$\\begin{aligned} (x-3)^2 = 0 &\\implies x = 3 \\\\ 4x-3 = 0 &\\implies x = \\frac{3}{4} \\end{aligned}$$"},{"type":"text","order":4,"title":"Calculate the y-coordinates","content":"Now we substitute our $x$-values back into the original curve equation $y = x(x-3)^3$ to find the corresponding $y$-coordinates.\n\n**For $x = 3$:**\n$$y = 3(3-3)^3 = 3(0)^3 = 0$$\nThis gives the stationary point **$(3, 0)$**.\n\n**For $x = \\frac{3}{4}$:**\n$$\\begin{aligned} y &= \\frac{3}{4}\\left(\\frac{3}{4}-3\\right)^3 \\\\ y &= \\frac{3}{4}\\left(-\\frac{9}{4}\\right)^3 \\\\ y &= \\frac{3}{4}\\left(-\\frac{729}{64}\\right) \\\\ y &= -\\frac{2187}{256} \\end{aligned}$$\nThis gives the stationary point **$\\left(\\frac{3}{4}, -\\frac{2187}{256}\\right)$**."},{"type":"text","order":5,"title":"Determine nature using sign table","content":"We will use the **First Derivative Test** by constructing a gradient sign table. We test the sign of $\\frac{dy}{dx} = (x-3)^2(4x-3)$ around our stationary points $x = 0.75$ and $x = 3$.\n\nLet's choose convenient test values:\n\n| $x$ | $x < 0.75$ (e.g., $0$) | $x = 0.75$ | $0.75 < x < 3$ (e.g., $1$) | $x = 3$ | $x > 3$ (e.g., $4$) |\n| :--- | :---: | :---: | :---: | :---: | :---: |\n| **Sign of $(x-3)^2$** | $+$ | $+$ | $+$ | $0$ | $+$ |\n| **Sign of $(4x-3)$** | $-$ | $0$ | $+$ | $+$ | $+$ |\n| **Sign of $\\frac{dy}{dx}$** | $-$ | $0$ | $+$ | $0$ | $+$ |\n| **Shape of curve** | $\\\\backslash$ | $\\_$ | $/$ | $\\_$ | $/$ |\n\nFrom the gradient patterns:\n* Around $x = \\frac{3}{4}$, the gradient changes from negative (decreasing) to positive (increasing), indicating a **local minimum**.\n* Around $x = 3$, the gradient stays positive (increasing) on both sides, indicating a **point of inflexion**."},{"type":"text","order":6,"title":"State the final conclusion","content":"We have successfully found and classified both stationary points:\n\n* **$\\left(\\frac{3}{4}, -\\frac{2187}{256}\\right)$** is a **local minimum**.\n* **$(3, 0)$** is a **point of inflexion**.\n\n### How marks are awarded:\n* **M1** for applying the Product Rule to get $\\frac{dy}{dx} = (x-3)^3 + 3x(x-3)^2$.\n* **M1** for factorising it to $(x-3)^2(4x-3)$.\n* **M1** for setting the derivative to $0$ and finding $x = 3$ and $x = \\frac{3}{4}$.\n* **A2** (1 mark each) for correctly calculating the $y$-coordinates: $(3,0)$ and $\\left(\\frac{3}{4}, -\\frac{2187}{256}\\right)$.\n* **M1** for setting up a gradient sign test around both stationary points.\n* **A2** (1 mark each) for correctly identifying the minimum and the point of inflexion."}]},{"id":"ed70ef43-3f43-4739-a177-9f1ea527098b:expansion_second_derivative","label":"Expansion and Second Derivative Test","steps":[{"type":"text","order":0,"title":"Expand the function","content":"To follow our chosen approach, we will first expand the expression for $y$ into a single polynomial. Let's start by expanding $(x-3)^3$ using the binomial expansion:\n\n$$(x-3)^3 = x^3 - 9x^2 + 27x - 27$$\n\nNow, multiply this result by $x$ to get the fully expanded polynomial form of the curve:\n\n$$y = x(x^3 - 9x^2 + 27x - 27) = x^4 - 9x^3 + 27x^2 - 27x$$"},{"type":"text","order":1,"title":"Find the first derivative","content":"Now we differentiate $y$ term-by-term with respect to $x$:\n\n$$\\frac{dy}{dx} = 4x^3 - 27x^2 + 54x - 27$$"},{"type":"text","order":2,"title":"Find the stationary points","content":"Stationary points occur when the gradient is zero, so we set $\\frac{dy}{dx} = 0$:\n\n$$4x^3 - 27x^2 + 54x - 27 = 0$$\n\nWe can factorise this polynomial. Since we know $(x-3)$ is likely a factor from the original expression, we can check if $x=3$ is a root: \n$$4(3)^3 - 27(3)^2 + 54(3) - 27 = 108 - 243 + 162 - 27 = 0$$\n\nThus, $(x-3)$ is a factor. Dividing the cubic by $(x-3)$, or looking for a common factor, we can rewrite it as:\n\n$$(x-3)^2(4x-3) = 0$$\n\nThis gives two solutions for $x$:\n\n$$x = 3 \\quad \\text{and} \\quad x = \\frac{3}{4}$$"},{"type":"text","order":3,"title":"Calculate the y-coordinates","content":"Substitute each $x$-value back into the original curve equation $y = x(x-3)^3$:\n\n- For $x = 3$:\n $$y = 3(3-3)^3 = 0$$\n This gives the stationary point $\\mathbf{(3, 0)}$.\n\n- For $x = \\frac{3}{4}$:\n $$y = \\frac{3}{4}\\left(\\frac{3}{4}-3\\right)^3 = \\frac{3}{4}\\left(-\\frac{9}{4}\\right)^3 = \\frac{3}{4}\\left(-\\frac{729}{64}\\right) = -\\frac{2187}{256}$$\n This gives the stationary point $\\mathbf{\\left(\\frac{3}{4}, -\\frac{2187}{256}\\right)}$."},{"type":"text","order":4,"title":"Find the second derivative","content":"To determine the nature of these points, we find the second derivative, $\\frac{d^2y}{dx^2}$, by differentiating our first derivative expression:\n\n$$\\frac{dy}{dx} = 4x^3 - 27x^2 + 54x - 27$$\n\n$$\\frac{d^2y}{dx^2} = 12x^2 - 54x + 54$$"},{"type":"hyper_latex","block":{"nodes":[{"latex":"$$\\text{Substitute } \\htmlData{anchor=xval1}{\\color{@sky}{x = \\frac{3}{4}}} \\text{ into } \\frac{d^2y}{dx^2} = 12\\htmlData{anchor=xsub1}{\\color{@sky}{x}}^2 - 54\\htmlData{anchor=xsub2}{\\color{@sky}{x}} + 54:$"},{"latex":"$$$\\begin{aligned} \\frac{d^2y}{dx^2} &= 12\\left(\\color{@sky}{\\frac{3}{4}}\\right)^2 - 54\\left(\\color{@sky}{\\frac{3}{4}}\\right) + 54 \\\\ &= 12\\left(\\frac{9}{16}\\right) - \\frac{81}{2} + 54 \\\\ &= \\frac{27}{4} - \\frac{162}{4} + \\frac{216}{4} \\\\ &= \\htmlData{anchor=d2yval}{\\color{@green}{\\frac{81}{4}}} > 0 \\end{aligned}$$"}],"highlights":[{"color":"sky","style":"text-color","anchor":"xval1"},{"color":"green","style":"box","anchor":"d2yval"}],"connections":[{"to":"xsub1","from":"xval1","color":"sky","label":"substitute","style":"solid"}]},"order":5,"title":"Test the first point","description":"Since the second derivative is positive at $x = \\frac{3}{4}$, this point is a local minimum."},{"type":"text","order":6,"title":"Test the second point","content":"Substitute $x = 3$ into the second derivative:\n\n$$\\frac{d^2y}{dx^2} = 12(3)^2 - 54(3) + 54 = 108 - 162 + 54 = 0$$\n\nSince $\\frac{d^2y}{dx^2} = 0$, the second derivative test is inconclusive on its own. We must check for a change in the sign of concavity (the sign of $\\frac{d^2y}{dx^2}$) just to the left and right of $x = 3$:\n\n- Let's test a point slightly to the left, $x = 2.9$:\n $$\\frac{d^2y}{dx^2} = 12(2.9)^2 - 54(2.9) + 54 = 100.92 - 156.6 + 54 = -1.68 < 0 \\quad \\text{(concave down)}$$\n- Let's test a point slightly to the right, $x = 3.1$:\n $$\\frac{d^2y}{dx^2} = 12(3.1)^2 - 54(3.1) + 54 = 115.32 - 167.4 + 54 = 1.92 > 0 \\quad \\text{(concave up)}$$\n\nBecause the second derivative changes sign from negative to positive through $x = 3$, the point $(3, 0)$ is a **point of inflexion**."},{"type":"text","order":7,"title":"Summarise the results","content":"Our final classifications for the stationary points are:\n- $\\mathbf{\\left(\\frac{3}{4}, -\\frac{2187}{256}\\right)}$ is a **local minimum**.\n- $\\mathbf{(3, 0)}$ is a **point of inflexion**."}]}],"generatedAt":"2026-06-09T05:04:29.435Z","modelVersion":"gemini-3.5-flash","explanationVersion":"v2"},"firstAttemptFullMarks":null,"walkthroughVideo":null},{"id":"01425972-1ed3-4db5-950c-3ff2baeadef5","revisionId":"dfadfbcb-6c26-4b04-92d4-7139eb00e862","questionNumber":null,"questionType":"LA","specification":"","difficulty":null,"difficultyBand":"hard","tier":null,"totalMarks":4,"sourceTitle":null,"calculatorUse":null,"isCalculation":true,"parts":[{"id":"a5dadffa-d7fe-4aa7-8ce7-88a82f4cab4a","label":null,"content":"The radius of a circle increases at $0.4\\,\\text{cm s}^{-1}$. Find the rate at which its area increases when the radius is $6$ cm.","markscheme":"- $A=\\pi r^2$, so $\\frac{dA}{dr}=2\\pi r$ M1\n- $\\frac{dA}{dt}=\\frac{dA}{dr}\\frac{dr}{dt}$ M1\n- $\\frac{dA}{dt}=2\\pi(6)(0.4)$ A1\n- $4.8\\pi\\,\\text{cm}^2\\text{s}^{-1}$ 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Differentiation","topicId":"02df946e-8a41-4062-ae8d-11af60f06e7e","topicSidebar":"$L3c","topicSlug":"differentiation","topicTitle":"1.6 Differentiation"}]}],"$L3d","$L3e"]}]

1.6 Differentiation

Question 2

1.6 Differentiation Questions

1.6 Differentiation

Question 2

1.6 Differentiation Questions