On the same diagram, sketch the graph of \({f^{ - 1}}\) .

Write down the range of \({f^{ - 1}}\) .

Find \({f^{ - 1}}(x)\) .

     A1A1A1     N3

Note: Award A1 for approximately correct (reflected) shape, A1 for right end point in circle, A1 for through \((1{\text{, }}0)\) .

\(0 \le y \le 3.5\)     A1     N1

[1 mark]

interchanging x and y (seen anywhere)     M1

e.g. \(x = {e^{0.5y}}\)

evidence of changing to log form     A1

e.g. \(\ln x = 0.5y\) , \(\ln x = \ln {{\rm{e}}^{0.5y}}\) (any base), \(\ln x = 0.5y\ln {\rm{e}}\) (any base)

\({f^{ - 1}}(x) = 2\ln x\)     A1     N1

[3 marks]

There were a large number of candidates who were unaware of the geometric relationship between a function and its inverse. Those that had some idea of the shape of the graph often did not consider the specified domain. Many more students were able to use an analytical approach to finding the inverse of a function and had little problem using logarithms to solve for y. Candidates were clearly more comfortable with algebraic procedures than graphical interpretations.

There were a large number of candidates who were unaware of the geometric relationship between a function and its inverse. Those that had some idea of the shape of the graph often did not consider the specified domain. Many more students were able to use an analytical approach to finding the inverse of a function and had little problem using logarithms to solve for y. Candidates were clearly more comfortable with algebraic procedures than graphical interpretations.

There were a large number of candidates who were unaware of the geometric relationship between a function and its inverse. Those that had some idea of the shape of the graph often did not consider the specified domain. Many more students were able to use an analytical approach to finding the inverse of a function and had little problem using logarithms to solve for y. Candidates were clearly more comfortable with algebraic procedures than graphical interpretations.