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June 2012 p13 q11
764
The function g is such that \(g(x) = 8 - (x - 2)^2\), for \(k \leq x \leq 4\), where \(k\) is a constant.
(ii) State the smallest value of \(k\) for which \(g\) has an inverse.
For this value of \(k\),
(iii) find an expression for \(g^{-1}(x)\),
(iv) sketch, on the same diagram, the graphs of \(y = g(x)\) and \(y = g^{-1}(x)\).
Solution
(ii) For \(g(x) = 8 - (x - 2)^2\) to have an inverse, it must be one-to-one. The function is a downward-opening parabola with vertex at \(x = 2\). To ensure it is one-to-one, we restrict the domain to \(x \geq 2\). Therefore, the smallest value of \(k\) is 2.
(iii) To find \(g^{-1}(x)\), start with \(y = 8 - (x - 2)^2\).
Rearrange to make \(x\) the subject:
\((x - 2)^2 = 8 - y\)
\(x - 2 = \pm \sqrt{8 - y}\)
Since \(x \geq 2\), we take the positive root:
\(x = 2 + \sqrt{8 - y}\)
Thus, \(g^{-1}(x) = 2 + \sqrt{8 - x}\).
(iv) The sketch should show the graph of \(y = g(x)\) as a downward-opening parabola starting at \(x = 2\) and ending at \(x = 4\). The graph of \(y = g^{-1}(x)\) should be the reflection of \(y = g(x)\) across the line \(y = x\), starting at \(y = 2\) and ending at \(y = 4\).
