1. **State the problem:** We have two functions defined as $f(x) = 1 - 2x$ and $g(x) = x^2 - 3$. (a) Determine whether $g^{-1}(x)$ is a function or not and give a reason. (b) Find: (i) $h(x)$ if $f^{-1}(h(x)) = g(x)$. (ii) The values of $x$ if $g(f^{-1}(x)) = 1$. 2. **Recall important concepts:** - The inverse of a function $f$, denoted $f^{-1}$, exists only if $f$ is one-to-one (bijective). - To find $f^{-1}$, solve $y = f(x)$ for $x$ in terms of $y$. - Composition of functions: $(g \, o \, f)(x) = g(f(x))$. 3. **Part (a): Is $g^{-1}(x)$ a function?** - $g(x) = x^2 - 3$ is a quadratic function. - Quadratic functions are not one-to-one over all real numbers because they fail the horizontal line test. - Therefore, $g^{-1}(x)$ is not a function unless the domain of $g$ is restricted. 4. **Part (b)(i): Find $h(x)$ if $f^{-1}(h(x)) = g(x)$** - First, find $f^{-1}(x)$: $$y = 1 - 2x \implies x = \frac{1 - y}{2}$$ Swap $x$ and $y$: $$f^{-1}(x) = \frac{1 - x}{2}$$ - Given $f^{-1}(h(x)) = g(x)$, substitute $f^{-1}$: $$\frac{1 - h(x)}{2} = g(x) = x^2 - 3$$ - Solve for $h(x)$: $$1 - h(x) = 2(x^2 - 3)$$ $$1 - h(x) = 2x^2 - 6$$ $$h(x) = 1 - (2x^2 - 6) = 1 - 2x^2 + 6 = 7 - 2x^2$$ 5. **Part (b)(ii): Find $x$ such that $g(f^{-1}(x)) = 1$** - Recall $f^{-1}(x) = \frac{1 - x}{2}$ - Compute $g(f^{-1}(x))$: $$g\left(f^{-1}(x)\right) = \left(\frac{1 - x}{2}\right)^2 - 3$$ - Set equal to 1: $$\left(\frac{1 - x}{2}\right)^2 - 3 = 1$$ - Simplify: $$\left(\frac{1 - x}{2}\right)^2 = 4$$ $$\frac{(1 - x)^2}{4} = 4$$ $$(1 - x)^2 = 16$$ - Take square root: $$1 - x = \pm 4$$ - Solve for $x$: - If $1 - x = 4$, then $x = -3$ - If $1 - x = -4$, then $x = 5$ **Final answers:** - (a) $g^{-1}(x)$ is not a function because $g$ is not one-to-one over all real numbers. - (b)(i) $h(x) = 7 - 2x^2$ - (b)(ii) $x = -3$ or $x = 5$