The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto (x-4)^2 + 1$ for \(x \in \m

Question

The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto (x-4)^2 + 1$ for $x \in \mathbb{R}, x > 4$. Sketch the graph of $y = \mathrm{f}(x)$. Your sketch should indicate the position of the graph in relation to the origin. [2] Find $\mathrm{f}^{-1}(x)$, stating the domain of $\mathrm{f}^{-1}$.[3] On the same diagram as in part (i), sketch the graph of $y = \mathrm{f}^{-1}(x)$.[1] Write down the equation of the line in which the graph of $y = \mathrm{f}(x)$ must be reflected in order to obtain the graph of $y = \mathrm{f}^{-1}(x)$, and hence find the exact solution of the equation $\mathrm{f}(x) = \mathrm{f}^{-1}(x)$.[4] The function $\mathrm{g}$ is defined by $\mathrm{g} : x \mapsto 3 + \frac{2}{2x-1}$ for $x \in \mathbb{R}, x \neq \frac{1}{2}$. Find the range of $\mathrm{gf}$.[2]

Accepted Answer

(i) See graph. (ii) $\mathrm f^{-1}(x)=4+\sqrt{x-1}$, domain $(1,\infty)$ (iii) See graph. (iv) $y=x$; $x=\dfrac{9+\sqrt{13}}2$ (v) $R_{\mathrm{gf}}=(3,5)$

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