}, -\\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"},"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 $\\frac{dy}{dx}$** | $2(-2)^2(-1) = -8 \\implies \\boldsymbol{-}$ | $0$ | $2(-1)^2(1) = 2 \\implies \\boldsymbol{+}$ | $0$ | $2(1)^2(5) = 10 \\implies \\boldsymbol{+}$

1.7 Differentiation

Question 1

1.7 Differentiation Questions

1.7 Differentiation

Question 1

1.7 Differentiation Questions