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 \(\sqrt {25} = \pm 5\).