Question4. Given $f(x)=e^{x}$ and $g(x)=|x-3|$. (a) Show that $(f \circ g)(x)=\left\{\begin{array}{ll}e^{x-3}, & x \geq 3 \\ e^{-(x-3)}, & x<3\end{array}\right.$. (b) Determine $(f \circ g)^{-1}(x)$, for $x \geq 3$. [3 marks] [4 marks] (c) Find the function $h(x)$ for $x>\frac{1}{3}$, given that $(h \circ f)(x)=\frac{2 e^{x}}{1-3 e^{x}}$. Hence, show that $h(x)$ is a one to one function.

Studdy Solution
Given $(h \circ f)(x) = \frac{2e^x}{1-3e^x}$, we need to find $h(x)$.
Let $u = e^x$, then $(h \circ f)(x) = h(u) = \frac{2u}{1-3u}$.
Thus, $h(x) = \frac{2x}{1-3x}$.
To show that $h(x)$ is one-to-one, we need to show that it is strictly increasing or decreasing. We can do this by finding the derivative $h'(x)$ and checking its sign:
$$h(x) = \frac{2x}{1-3x}$$
Using the quotient rule:
$$h'(x) = \frac{(1-3x)(2) - (2x)(-3)}{(1-3x)^2}$$
Simplifying:
$$h'(x) = \frac{2 - 6x + 6x}{(1-3x)^2} = \frac{2}{(1-3x)^2}$$
Since the denominator $(1-3x)^2$ is always positive for $x > \frac{1}{3}$, $h'(x) > 0$. Therefore, $h(x)$ is strictly increasing and thus one-to-one.
The solutions are: (a) $(f \circ g)(x) = \begin{cases} e^{x-3}, & x \geq 3 \\ e^{-(x-3)}, & x < 3 \end{cases}$ (b) $(f \circ g)^{-1}(x) = \ln(x) + 3$ (c) $h(x) = \frac{2x}{1-3x}$, and $h(x)$ is one-to-one.