Find ${f^{ - 1}}(2)$ .

Let $g$ be a function such that ${g^{ - 1}}$ exists for all real numbers. Given that $g(30) = 3$ , find $(f \circ {g^{ - 1}})(3)$  .

METHOD 1

attempt to set up equation     (M1)

eg   $2 = \sqrt {y - 5}$ , $2 = \sqrt {x - 5}$

correct working     (A1)

eg   $4 = y - 5$ , $x = {2^2} + 5$

${f^{ - 1}}(2) = 9$     A1     N2

METHOD 2

interchanging $x$ and $y$ (seen anywhere)     (M1)

eg   $x = \sqrt {y - 5}$

correct working     (A1)

eg   ${x^2} = y - 5$ , $y = {x^2} + 5$

${f^{ - 1}}(2) = 9$     A1     N2

[3 marks]

recognizing ${g^{ - 1}}(3) = 30$     (M1)

eg   $f(30)$

correct working     (A1)

eg   $(f \circ {g^{ - 1}})(3) = \sqrt {30 - 5}$ , $\sqrt {25}$

$(f \circ {g^{ - 1}})(3) = 5$     A1     N2

Note: Award A0 for multiple values, eg $\pm 5$ .

[3 marks]

Candidates often found an inverse function in which to substitute the value of $2$. Some astute candidates set the function equal to $2$ and solved for $x$. Occasionally a candidate misunderstood the notation as asking for a derivative, or used $\frac{1}{{f(x)}}$ .

For part (b), many candidates recognized that if $g(30) = 3$ then ${g^{ - 1}}(3) = 30$ , and typically completed the question successfully. Occasionally, however, a candidate incorrectly answered $\sqrt {25} = \pm 5$.