0606 P12 - Jun 2020 - Q5 - 9 marks
8112
The function \(f\) is defined by
\(f:x\mapsto(2x+3)^2,\qquad x\gt0.\)
(a) State the range of \(f\).
(b) Explain why \(f\) has an inverse.
(c) Find \(f^{-1}\).
(d) State the domain of \(f^{-1}\).
(e) Given that
\(g:x\mapsto\ln(x+4),\qquad x\gt0,\)
find the exact solution of \(fg(x)=49\).
Solution
Answer: range \(f\gt9\); \(f^{-1}(x)=\dfrac{\sqrt x-3}{2}\), domain \(x\gt9\); \(x=e^2-4\).
(a) Since \(x\gt0\),
\(2x+3\gt3.\)
Therefore
\((2x+3)^2\gt9.\)
So the range is
\(f\gt9.\)
(b) On the restricted domain \(x\gt0\), the function \(f(x)=(2x+3)^2\) is one-to-one. Therefore \(f\) has an inverse.
(c) Let
\(y=(2x+3)^2.\)
Since \(x\gt0\), \(2x+3\gt0\), so
\(\sqrt y=2x+3.\)
Hence
\(x=\frac{\sqrt y-3}{2}.\)
Therefore
\(f^{-1}(x)=\frac{\sqrt x-3}{2}.\)
(d) The domain of \(f^{-1}\) is the range of \(f\), so
\(x\gt9.\)
(e) We need
\(f(g(x))=49.\)
Since \(g(x)=\ln(x+4)\),
\(\left(2\ln(x+4)+3\right)^2=49.\)
Because \(x\gt0\), we have \(\ln(x+4)\gt0\), so \(2\ln(x+4)+3\) is positive. Therefore
\(2\ln(x+4)+3=7.\)
Thus
\(\ln(x+4)=2.\)
So
\(x+4=e^2,\)
and hence
\(x=e^2-4.\)
