The function f is defined, for $x > \frac{3}{2}$, by $\text{f}(x) = \frac{e^{2x}}{1 - x}$. Explain clearly whether f is an increasing or a decreasing function.

1. Find the first derivative of $f(x)$ using the quotient rule.

$\text{f}'(x) = \frac{(1 - x)\frac{d}{dx}(e^{2x}) - e^{2x}\frac{d}{dx}(1 - x)}{(1 - x)^2}$
$\text{f}'(x) = \frac{(1 - x)(2e^{2x}) - e^{2x}(-1)}{(1 - x)^2}$

2. Simplify the derivative.

$\text{f}'(x) = \frac{2e^{2x} - 2xe^{2x} + e^{2x}}{(1 - x)^2}$
$\text{f}'(x) = \frac{e^{2x}(3 - 2x)}{(1 - x)^2}$

3. Analyse the sign of the derivative for the given domain.

For $x > \frac{3}{2}$:

• $e^{2x} > 0$ for all real $x$.
• $(1 - x)^2 > 0$ since $x \neq 1$.
• $3 - 2x < 0$ because $2x > 3$.
  Therefore, $\text{f}'(x) = \frac{(+)(-)}{(+)} < 0$.

4. State the conclusion.

Since $\text{f}'(x) < 0$ for all $x > \frac{3}{2}$, $\text{f}$ is a decreasing function.