The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto \frac{1}{x^2 + 6x + 5}$, for&nbs

Question

The function $\mathrm{f}$ is defined by $\mathrm{f} : x \mapsto \frac{1}{x^2 + 6x + 5}$, for $x \ge -3$, $x \neq -1$. Find $\mathrm{f}^{-1}(x)$.[3] Find algebraically the range of $\mathrm{f}$.[4] The function $\mathrm{g}$ is defined by $\mathrm{g} : x \mapsto e^x$, for $x \in \mathbb{R}$. Find the exact range of $\mathrm{gf}$.[2]

Accepted Answer

(a) $\mathrm f^{-1}(x)=-3+\sqrt{4+\dfrac1x}$, with domain $\left(-\infty,-\dfrac14\right]\cup(0,\infty)$ (b) $R_{\mathrm f}=\left(-\infty,-\dfrac14\right]\cup(0,\infty)$ (c) $R_{\mathrm{gf}}=(0,e^{-1/4}]\cup(1,\infty)$

2025 ACJC P2 Q4

2025 ACJC P2 Q4

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