The function $\mathrm{f}$ is defined by $\mathrm{f}:x \mapsto \frac{1}{2}+e^{-x}$, \(x \in \math

Question

The function $\mathrm{f}$ is defined by $\mathrm{f}:x \mapsto \frac{1}{2}+e^{-x}$, $x \in \mathbb{R}$, $x \ge 0$. Sketch the graph of $y=\mathrm{f}(x)$, $y=\mathrm{f}^{-1}(x)$ and $y=\mathrm{f}^{-1}\mathrm{f}(x)$ on the same diagram, showing clearly their relationship.[3] The function $\mathrm{g}$ is defined by $\mathrm{g}:x \mapsto 1+4x-x^2$, $x \le 2$. Explain why $\mathrm{g}^{-1}$ exists.[1] Find $\mathrm{g}^{-1}(x)$ and state its domain.[3] Explain why the composite function $\mathrm{gf}$ exists.[2] Find the exact range of $\mathrm{gf}$.[2]

Accepted Answer

(a) See graph. (b) $\mathrm g^{-1}$ exists because $\mathrm g$ is one-to-one. (c) $\mathrm g^{-1}(x)=2-\sqrt{5-x}$, $x\le5$ (d) $R_{\mathrm f}=(\frac12,\frac32]\subseteq D_{\mathrm g}$ (e) $R_{\mathrm{gf}}=(\frac{11}{4},\frac{19}{4}]$

2025 TJC P2 Q3

2025 TJC P2 Q3

Video Solution:

Solution:

Solution locked