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Different inputs may share an output, but a single input cannot have multiple outputs."},{"text":"the same","correct":false,"explanation":"A function does not have to give the same output for every input. It must give exactly one output for each individual input."},{"text":"a random","correct":false,"explanation":"Functions are deterministic rules rather than random processes. Each permitted input must be assigned exactly one specific output, so the correct term is one."},{"text":"an unpredictable","correct":false,"explanation":"An output cannot be left unpredictable while still defining a function. The rule must associate every permitted input with exactly one determined output."}]},{"id":"10a8cb41-01c5-8362-641e-8a69c5e1b30a","cardType":"basic","points":1,"front":"How do you evaluate $h(4)$ for the function $h(x) = 2x^2 - 6$?","back":"Substitute 4 for $x$ in the formula: $2(4)^2 - 6$.","mcqOptions":[{"text":"Substitute 4 for $x$ in the formula: $2(4)^2 - 6$.","correct":true,"explanation":"Evaluating $h(4)$ means replacing every $x$ in $h(x)=2x^2-6$ with 4, giving $2(4)^2-6$ before calculating the result."},{"text":"Substitute $-4$ for $x$ in the formula: $2(-4)^2 - 6$.","correct":false,"explanation":"This substitutes $-4$, so it evaluates $h(-4)$ rather than $h(4)$. The required substitution is $2(4)^2-6$."},{"text":"Substitute 4 for $x$ in the formula: $(2 \\times 4)^2 - 6$.","correct":false,"explanation":"This incorrectly squares the product $2\\times4$ instead of squaring only the input. The coefficient remains outside the square, giving $2(4)^2-6$."},{"text":"Substitute 2 for $x$ in the formula: $2(2)^2 - 6$.","correct":false,"explanation":"This substitutes the coefficient 2 instead of the requested input 4. To evaluate $h(4)$, replace $x$ with 4 to obtain $2(4)^2-6$."}]},{"id":"1ed87258-05eb-57ea-c359-194d5903b0c7","cardType":"cloze","points":null,"front":"When substituting negative numbers into a function, always use [ ] to avoid calculation errors.","back":"When substituting negative numbers into a function, always use brackets to avoid calculation errors.","mcqOptions":[{"text":"brackets","correct":true,"explanation":"Brackets keep the negative sign attached to the substituted value. For example, $(-3)^2=9$, whereas writing $-3^2$ is interpreted as $-(3^2)=-9$."},{"text":"a calculator","correct":false,"explanation":"A calculator does not prevent an incorrectly entered negative substitution. 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Depending on the function, there may be one, several, or no such inputs."},{"text":"Find the output values that result from an input of 10.","correct":false,"explanation":"This reverses the roles of input and output. The output is already specified as 10, so the task is to find the corresponding input values."},{"text":"Find the output value that results from an input of 10.","correct":false,"explanation":"Setting the input to 10 would evaluate $f(10)$, not solve $f(x)=10$. Solving requires finding inputs whose output is equal to 10."},{"text":"Find the input values that result in an output greater than 10.","correct":false,"explanation":"The equation asks for inputs producing exactly 10, not outputs greater than 10. The required values satisfy the equality $f(x)=10$."}]},{"id":"02d4464e-6c7a-15a6-82cb-2d81cea3f84f","cardType":"basic","points":1,"front":"If $x^2 = 9$, why are there two possible solutions for $x$?","back":"Both positive and negative values give a positive result when squared ($3^2 = 9$ and $(-3)^2 = 9$).","mcqOptions":[{"text":"Both positive and negative values give a positive result when squared ($3^2 = 9$ and $(-3)^2 = 9$).","correct":true,"explanation":"Squaring removes the distinction between opposite signs because the product of two negative numbers is positive. Therefore both $3$ and $-3$ satisfy $x^2=9$."},{"text":"Only the positive value gives a positive result when squared ($3^2 = 9$, so $x = 3$ is the only solution).","correct":false,"explanation":"This keeps only the principal square root and overlooks the second solution. Since $(-3)^2$ also equals 9, both $x=3$ and $x=-3$ work."},{"text":"The positive value gives 9 and the negative value gives $-9$ when squared ($3^2 = 9$ and $(-3)^2 = -9$).","correct":false,"explanation":"Squaring a negative number produces a positive result because two negative factors multiply to give a positive product. Thus $(-3)^2=9$, not $-9$."},{"text":"The positive and negative values cancel when squared, so the only possible solution is $x = 0$.","correct":false,"explanation":"Opposite values do not cancel when each is squared separately. Both $3^2$ and $(-3)^2$ equal 9, giving the two solutions $x=3$ and $x=-3$."}]},{"id":"d728b838-f1ba-34ee-83e3-c52dbe8cdd4a","cardType":"basic","points":1,"front":"What is the relationship between the outputs of $f(x)$ and $f^{-1}(x)$?","back":"The inverse function $f^{-1}$ undoes the function $f$ (if $f(a) = b$, then $f^{-1}(b) = a$).","mcqOptions":[{"text":"The inverse function $f^{-1}$ undoes the function $f$ (if $f(a) = b$, then $f^{-1}(b) = a$).","correct":true,"explanation":"An inverse reverses the original mapping: if $f$ sends $a$ to $b$, then $f^{-1}$ sends $b$ back to $a$. Their inputs and outputs exchange roles."},{"text":"The inverse function $f^{-1}$ reciprocates the function $f$ (if $f(a) = b$, then $f^{-1}(a) = \\frac{1}{b}$).","correct":false,"explanation":"The superscript $-1$ denotes an inverse function, not a reciprocal. If $f(a)=b$, the inverse maps $b$ back to $a$ rather than calculating $1/b$."},{"text":"The inverse function $f^{-1}$ repeats the function $f$ (if $f(a) = b$, then $f^{-1}(a) = b$).","correct":false,"explanation":"Repeating the same mapping describes applying $f$ again, not applying its inverse. The inverse must reverse the mapping, so $f^{-1}(b)=a$."},{"text":"The inverse function $f^{-1}$ negates the function $f$ (if $f(a) = b$, then $f^{-1}(a) = -b$).","correct":false,"explanation":"An inverse does not generally change an output’s sign. It reverses the input-output mapping, meaning that $f(a)=b$ leads to $f^{-1}(b)=a$."}]},{"id":"5dc7bbc9-4be2-f776-6fc7-3ec01929f8af","cardType":"basic","points":3,"front":"Outline the three steps to find an inverse function $f^{-1}(x)$.","back":"

Set $y = f(x)$
Swap $x$ and $y$
Rearrange the equation to make $y$ the subject

","mcqOptions":[{"text":"

Set $y = f(x)$
Swap $x$ and $y$
Rearrange the equation to make $y$ the subject

","correct":true,"explanation":"Writing the function as $y$, exchanging input and output variables, and then solving for $y$ reverses the original mapping and produces the inverse function."},{"text":"

Set $y = f(x)$
Keep $x$ and $y$ unchanged
Rearrange the equation to make $y$ the subject

","correct":false,"explanation":"Leaving $x$ and $y$ unchanged does not reverse the input-output relationship, so it merely rearranges the original function. They must be swapped before solving for $y$."},{"text":"

Set $y = f(x)$
Swap $x$ and $y$
Rearrange the equation to make $x$ the subject

","correct":false,"explanation":"After swapping $x$ and $y$, solving for $x$ moves back towards the original function. To express the inverse in function form, rearrange to make $y$ the subject."},{"text":"

Set $y = f(x)$
Negate $x$ and $y$
Rearrange the equation to make $y$ the subject

","correct":false,"explanation":"Negating both variables reflects signs rather than reversing the input-output mapping. The correct process swaps $x$ and $y$, then rearranges to make $y$ the subject."}]},{"id":"5da4decf-4bc4-629e-756d-8970e38fed12","cardType":"cloze","points":null,"front":"The notation $f^{-1}(x)$ refers to the [ ], not the reciprocal ($\\frac{1}{f(x)}$).","back":"The notation $f^{-1}(x)$ refers to the inverse function, not the reciprocal ($\\frac{1}{f(x)}$).","mcqOptions":[{"text":"inverse function","correct":true,"explanation":"The superscript $-1$ indicates the function that reverses the mapping performed by $f$. It does not mean taking the reciprocal of the function’s output."},{"text":"reciprocal function","correct":false,"explanation":"Treating the exponent as an ordinary negative power causes this misconception. The reciprocal is written $\\frac{1}{f(x)}$, while $f^{-1}(x)$ denotes the inverse function."},{"text":"negative function","correct":false,"explanation":"A negative function would be written $-f(x)$ and changes the signs of outputs. The notation $f^{-1}(x)$ instead represents the inverse function."},{"text":"derivative function","correct":false,"explanation":"A derivative represents the rate of change and is usually written $f'(x)$. The notation $f^{-1}(x)$ refers specifically to the inverse function."}]},{"id":"f0d7c6d1-5481-84fe-fdc4-5601db1f2171","cardType":"cloze","points":null,"front":"The notation $f^{-1}(x)$ refers to the inverse function, not the [ ] ($\\frac{1}{f(x)}$).","back":"The notation $f^{-1}(x)$ refers to the inverse function, not the reciprocal ($\\frac{1}{f(x)}$).","mcqOptions":[{"text":"reciprocal","correct":true,"explanation":"The exponent $-1$ attached to a function’s name denotes its inverse, whereas the reciprocal is $\\frac{1}{f(x)}$. These are different operations."},{"text":"negative","correct":false,"explanation":"A negative function is $-f(x)$, which changes the sign of every output. The expression $\\frac{1}{f(x)}$ is the reciprocal, not the negative."},{"text":"derivative","correct":false,"explanation":"A derivative describes the rate of change and is usually written $f'(x)$. The expression $\\frac{1}{f(x)}$ is instead called the reciprocal."},{"text":"square","correct":false,"explanation":"The square of the function would be $[f(x)]^2$, not $\\frac{1}{f(x)}$. Dividing $1$ by the function gives its reciprocal."}]},{"id":"ddd15fb6-0fa4-d3f2-bcd0-9b1c88c0d314","cardType":"basic","points":2,"front":"What is the inverse of $r(x) = 4x - 3$?","back":"$$r^{-1}(x) = \\frac{x+3}{4}$","mcqOptions":[{"text":"$$r^{-1}(x) = \\frac{x+3}{4}$","correct":true,"explanation":"Writing $y=4x-3$, swapping $x$ and $y$, and rearranging gives $x=4y-3$, so $y=\\frac{x+3}{4}$."},{"text":"$$r^{-1}(x) = \\frac{x-3}{4}$","correct":false,"explanation":"This keeps the subtraction rather than reversing it. After swapping variables, adding $3$ to both sides gives $x+3=4y$, so the numerator is $x+3$."},{"text":"$$r^{-1}(x) = 4x+3$","correct":false,"explanation":"This reverses the subtraction but not the multiplication. Both operations must be undone in reverse order, giving $r^{-1}(x)=\\frac{x+3}{4}$."},{"text":"$$r^{-1}(x) = \\frac{x+4}{3}$","correct":false,"explanation":"This mistakenly swaps the constants rather than reversing the operations. Add $3$ and then divide by $4$, giving $r^{-1}(x)=\\frac{x+3}{4}$."}]},{"id":"bea55fb9-f61b-f1b3-1913-d39a6a2ed231","cardType":"basic","points":1,"front":"In the composite function $gf(x)$, which function is applied to $x$ first?","back":"Function $f$ (the function nearest to $x$).","mcqOptions":[{"text":"Function $f$ (the function nearest to $x$).","correct":true,"explanation":"The notation $gf(x)$ means $g(f(x))$, so $x$ is first input into $f$. The resulting output is then used as the input to $g$."},{"text":"Function $g$ (the function furthest from $x$).","correct":false,"explanation":"Reading the letters from left to right causes this mistake. Since $gf(x)=g(f(x))$, function $f$, which is nearest to $x$, is applied first."},{"text":"Function $g$ (the function nearest to $x$).","correct":false,"explanation":"Although $g$ appears first when reading normally, it is not nearest to the input. The notation expands to $g(f(x))$, so $f$ acts first."},{"text":"Function $f$ (the function furthest from $x$).","correct":false,"explanation":"This identifies the correct letter but gives the wrong positional rule. In $gf(x)$, $f$ is nearest to $x$, which is why it is applied first."}]},{"id":"9591eda1-b87b-cc51-c63f-79c9fcf4caee","cardType":"cloze","points":null,"front":"Composite functions like $gf(x)$ are read from [ ].","back":"Composite functions like $gf(x)$ are read from right to left.","mcqOptions":[{"text":"right to left","correct":true,"explanation":"The notation $gf(x)$ represents $g(f(x))$. Therefore, the right-hand function $f$ is applied first, followed by the left-hand function $g$."},{"text":"left to right","correct":false,"explanation":"Ordinary reading order is tempting, but it reverses the composition. Since $gf(x)=g(f(x))$, the right-hand function $f$ must be applied before $g$."},{"text":"outside to inside","correct":false,"explanation":"Applying the outside function first reverses the required order. In $g(f(x))$, the inside function $f$ acts on $x$ before the outside function $g$."},{"text":"either way round","correct":false,"explanation":"Function composition is generally order-dependent, so the direction cannot be chosen freely. The notation $gf(x)$ must be read from right to left as $g(f(x))$."}]},{"id":"531c81f1-b133-2905-2b0a-5d18e832a716","cardType":"basic","points":1,"front":"Why is $fg(x)$ not the same as $f(x) \\times g(x)$?","back":"Composite notation indicates a chain of operations, not multiplication of the two formulas.","mcqOptions":[{"text":"Composite notation indicates a chain of operations, not multiplication of the two formulas.","correct":true,"explanation":"The notation $fg(x)$ means $f(g(x))$: the output of $g$ becomes the input of $f$. It does not mean the product $f(x)g(x)$."},{"text":"Composite notation indicates a sum of operations, not multiplication of the two formulas.","correct":false,"explanation":"Composition does not add the functions; that would be written $f(x)+g(x)$. Instead, $fg(x)$ means substituting $g(x)$ into $f$, forming a chain of operations."},{"text":"Composite notation indicates a difference of operations, not multiplication of the two formulas.","correct":false,"explanation":"Composition is not subtraction, which would be written $f(x)-g(x)$. The notation $fg(x)$ means $f(g(x))$, where one function acts on the other’s output."},{"text":"Composite notation indicates a division of operations, not multiplication of the two formulas.","correct":false,"explanation":"Composition does not divide the function formulas; a quotient would require a fraction. Instead, $fg(x)=f(g(x))$, so the functions are applied one after the other."}]},{"id":"46e7afb8-a60b-008f-79df-4b94d4fb69dc","cardType":"basic","points":2,"front":"Given $f(x) = 2x + 1$ and $g(x) = x^2$, calculate $fg(x)$.","back":"$$2x^2 + 1$","mcqOptions":[{"text":"$$2x^2 + 1$","correct":true,"explanation":"The composition $fg(x)$ means $f(g(x))$. Substituting $g(x)=x^2$ into $f(x)=2x+1$ gives $f(x^2)=2x^2+1$."},{"text":"$$(2x + 1)^2$","correct":false,"explanation":"This calculates $g(f(x))$, so the functions have been applied in the reverse order. For $fg(x)=f(g(x))$, the result is $2x^2+1$."},{"text":"$$4x^2 + 1$","correct":false,"explanation":"This incorrectly squares the coefficient $2$ when substituting $x^2$ into $f$. Only $x$ is replaced, so $2(x^2)+1=2x^2+1$."},{"text":"$$2x^2 + 2$","correct":false,"explanation":"This mistakenly changes the constant term when substituting into the function. The $+1$ remains unchanged, so $f(x^2)=2x^2+1$."}]},{"id":"8d097af6-10cd-1143-842a-e913a8e46767","cardType":"basic","points":2,"front":"Given $f(x) = 2x + 1$ and $g(x) = x^2$, calculate $gf(x)$.","back":"$$(2x + 1)^2$","mcqOptions":[{"text":"$$(2x + 1)^2$","correct":true,"explanation":"The composition $gf(x)$ means $g(f(x))$. Since $g$ squares its input, substituting $f(x)=2x+1$ gives $g(2x+1)=(2x+1)^2$."},{"text":"$$2x^2 + 1$","correct":false,"explanation":"This calculates $f(g(x))$, applying the functions in the reverse order. For $gf(x)=g(f(x))$, the whole expression $2x+1$ must be squared."},{"text":"$$4x^2 + 1$","correct":false,"explanation":"This squares the two terms separately and omits the middle term. The unexpanded answer is $(2x+1)^2$, which expands to $4x^2+4x+1$."},{"text":"$$4x^2 + 2x + 1$","correct":false,"explanation":"This underestimates the cross term when expanding the square. Since $2(2x)(1)=4x$, $(2x+1)^2$ expands to $4x^2+4x+1$."}]},{"id":"fe61a890-178b-39b3-f76d-c4efa3f06077","cardType":"cloze","points":null,"front":"When finding $f(x-3)$, you must substitute the [ ] $(x-3)$ wherever $x$ appears in the original function.","back":"When finding $f(x-3)$, you must substitute the entire expression $(x-3)$ wherever $x$ appears in the original function.","mcqOptions":[{"text":"entire expression","correct":true,"explanation":"The input to the function is now $x-3$, so every occurrence of the original input $x$ must be replaced by the bracketed expression $(x-3)$."},{"text":"variable only","correct":false,"explanation":"Substituting only the variable can lose the subtraction or apply it to just one occurrence. The whole input $(x-3)$ must replace every $x$ in the function."},{"text":"constant term","correct":false,"explanation":"The constant term in the original formula is not the new input. To find $f(x-3)$, replace each occurrence of $x$ with the entire expression $(x-3)$."},{"text":"final term","correct":false,"explanation":"Changing only the final term confuses substitution with altering part of the formula. The entire expression $(x-3)$ must replace every occurrence of the original variable $x$."}]},{"id":"4e322548-1746-ee13-79b9-8c02ad683d0c","cardType":"basic","points":2,"front":"How do you solve $(x - 3)^2 = 9$ without expanding the brackets?","back":"Square root both sides to get $x - 3 = 3$ and $x - 3 = -3$, then solve for $x$.","mcqOptions":[{"text":"Square root both sides to get $x - 3 = 3$ and $x - 3 = -3$, then solve for $x$.","correct":true,"explanation":"A square equal to $9$ can have a base of either $3$ or $-3$. Solving both resulting equations gives the two solutions $x = 6$ and $x = 0$."},{"text":"Square root both sides to get $x - 3 = 3$ only, then solve for $x$.","correct":false,"explanation":"Taking only the positive square root misses one solution. Since both $3^2$ and $(-3)^2$ equal $9$, you must solve $x - 3 = 3$ and $x - 3 = -3$."},{"text":"Square root both sides to get $x - 3 = 9$ and $x - 3 = -9$, then solve for $x$.","correct":false,"explanation":"This treats the square root of $9$ as $9$. The square root is $3$, so the required equations are $x - 3 = 3$ and $x - 3 = -3$."},{"text":"Square root both sides to get $x + 3 = 3$ and $x + 3 = -3$, then solve for $x$.","correct":false,"explanation":"Taking square roots does not change $x - 3$ into $x + 3$. The bracketed expression remains $x - 3$, which must equal either $3$ or $-3$."}]},{"id":"976d77b3-ae57-e06f-1ff4-2fb7cb332288","cardType":"basic","points":2,"front":"Given $a(x) = 3x - 4$ and $b(x) = kx^2$, how do you find $ab(2)$?","back":"

Calculate $b(2) = 4k$
Substitute $4k$ into function $a$: $3(4k) - 4$

","mcqOptions":[{"text":"

Calculate $b(2) = 4k$
Substitute $4k$ into function $a$: $3(4k) - 4$

","correct":true,"explanation":"The notation $ab(2)$ means $a(b(2))$, so evaluate the inner function first. Since $b(2)=k(2)^2=4k$, substituting into $a$ gives $3(4k)-4$."},{"text":"

Calculate $a(2) = 2$
Substitute $2$ into function $b$: $k(2)^2$

","correct":false,"explanation":"This reverses the order of the composite functions and calculates $b(a(2))$. For $ab(2)=a(b(2))$, calculate $b(2)=4k$ first, then substitute $4k$ into $a$."},{"text":"

Calculate $b(2) = 2k$
Substitute $2k$ into function $a$: $3(2k) - 4$

","correct":false,"explanation":"This forgets to square the input in $b(x)=kx^2$. Substituting $x=2$ gives $b(2)=k(2^2)=4k$, which is then substituted into $a$."},{"text":"

Calculate $b(2) = 4k$
Substitute $4k$ into function $a$: $3(4k - 4)$

","correct":false,"explanation":"This incorrectly places the subtraction inside the multiplication by $3$. Since $a(x)=3x-4$, substituting $x=4k$ gives $3(4k)-4$, not $3(4k-4)$."}]},{"id":"324f6ab3-2bdb-1016-8bc9-2d25825dfb75","cardType":"cloze","points":null,"front":"In the absence of a restricted domain, the inverse of a quadratic like $f(x) = x^2 - 9$ is often given as the [ ] square root.","back":"In the absence of a restricted domain, the inverse of a quadratic like $f(x) = x^2 - 9$ is often given as the non-negative square root.","mcqOptions":[{"text":"non-negative","correct":true,"explanation":"The principal square root is defined to produce values greater than or equal to zero. Therefore, the inverse branch of the quadratic is taken as the non-negative square root."},{"text":"positive","correct":false,"explanation":"The principal square root can equal zero, so describing it as positive is too restrictive. The inverse uses the non-negative square root, allowing outputs greater than or equal to zero."},{"text":"negative","correct":false,"explanation":"The negative square-root branch would produce values less than or equal to zero. By convention, the inverse is usually given using the principal, non-negative square root."},{"text":"non-positive","correct":false,"explanation":"Non-positive includes zero and negative values, which describes the lower square-root branch. The usual principal square root is non-negative, giving values greater than or equal to zero."}]},{"id":"2872f99f-e7f2-f894-6114-0689853a7276","cardType":"basic","points":1,"front":"What does the statement $f(x) = g(x)$ require you to do algebraically?","back":"Set the formula for function $f$ equal to the formula for function $g$ and solve for $x$.","mcqOptions":[{"text":"Set the formula for function $f$ equal to the formula for function $g$ and solve for $x$.","correct":true,"explanation":"The equation states that both functions have the same output for particular input values. Equating their algebraic formulas and solving identifies every value of $x$ where their graphs intersect."},{"text":"Substitute the formula for function $g$ into the formula for function $f$ and solve for $x$.","correct":false,"explanation":"Substituting one function into another forms the composite function $f(g(x))$ rather than solving $f(x)=g(x)$. The two original formulas must instead be set equal."},{"text":"Set the formula for function $f$ equal to the negative formula for function $g$ and solve for $x$.","correct":false,"explanation":"Introducing a negative sign changes the equation to $f(x)=-g(x)$, which represents different output values. For $f(x)=g(x)$, equate the two formulas directly and solve for $x$."},{"text":"Set the inverse formula for function $f$ equal to the formula for function $g$ and solve for $x$.","correct":false,"explanation":"Using the inverse of $f$ changes the equation and is not required here. Replace $f(x)$ and $g(x)$ with their original formulas, set them equal, and solve for $x$."}]},{"id":"46b62442-7521-6019-e266-cb588971216a","cardType":"basic","points":2,"front":"If $f(x) = x^2 + 2$, what is the expression for $f(x-3)$?","back":"$$(x-3)^2 + 2$","mcqOptions":[{"text":"$$(x-3)^2 + 2$","correct":true,"explanation":"Replacing the input $x$ with $x-3$ means substituting $x-3$ everywhere $x$ appears in the function. Therefore, the squared term becomes $(x-3)^2$ while the constant remains $2$."},{"text":"$$x^2 - 3 + 2$","correct":false,"explanation":"This subtracts $3$ after squaring rather than replacing the function’s input. Substituting $x-3$ for $x$ makes the entire input squared, giving $(x-3)^2+2$."},{"text":"$$(x+3)^2 + 2$","correct":false,"explanation":"This reverses the sign of the substituted input. The expression requested is $f(x-3)$, so $x$ must be replaced by $x-3$, giving $(x-3)^2+2$."},{"text":"$$(x-3)^2 - 2$","correct":false,"explanation":"Substituting $x-3$ changes only occurrences of the input variable, not the separate constant term. The $+2$ remains unchanged, so the expression is $(x-3)^2+2$."}]}]}],"examCountdown":{"examDate":"2027-05-14","paperLabel":"Paper 1"},"heading":"Inverse and Composite Functions","title":"Inverse and Composite Functions Flashcards","topbar":{"backHref":"/gcse/maths/ocr/flashcards"},"topicSidebar":"$L37"}]}],"topicId":"2fedbcb1-c57b-4dd3-82bb-d0fe917efa00","walled":"$L38"}],"$L39","$L3a"]