# Is f(x)=(x^2-e^x)/(x-2) increasing or decreasing at x=-1?

Dec 11, 2015

Increasing.

#### Explanation:

If $f ' \left(- 1\right) < 0$, then $f \left(x\right)$ is decreasing at $x = - 1$.
If $f ' \left(- 1\right) > 0$, then $f \left(x\right)$ is increasing at $x = - 1$.

Find the first derivative.

Use the quotient rule.

$f ' \left(x\right) = \frac{\left(x - 2\right) \frac{d}{\mathrm{dx}} \left[{x}^{2} - {e}^{x}\right] - \left({x}^{2} - {e}^{x}\right) \frac{d}{\mathrm{dx}} \left[x - 2\right]}{x - 2} ^ 2$

Find each derivative separately.

$\frac{d}{\mathrm{dx}} \left[{x}^{2} - {e}^{x}\right] = 2 x - {e}^{x}$

$\frac{d}{\mathrm{dx}} \left[x - 2\right] = 1$

Plug them back in.

$f ' \left(x\right) = \frac{\left(x - 2\right) \left(2 x - {e}^{x}\right) - \left({x}^{2} - {e}^{x}\right)}{x - 2} ^ 2$

$f ' \left(x\right) = \frac{2 {x}^{2} - 4 x - x {e}^{x} + 2 {e}^{x} - {x}^{2} + {e}^{x}}{x - 2} ^ 2$

$f ' \left(x\right) = \frac{{x}^{2} - x {e}^{x} + 3 {e}^{x} - 4 x}{x - 2} ^ 2$

Find $f ' \left(- 1\right)$.

$f ' \left(- 1\right) = \frac{{\left(- 1\right)}^{2} - \left(- 1\right) {e}^{- 1} + 3 {e}^{- 1} - 4 \left(- 1\right)}{\left(- 1\right) - 2} ^ 2$

$f ' \left(- 1\right) = \frac{5 + \frac{4}{e}}{9}$

We could determine the exact value of this number, but it's clear that it will be positive. Because of this, we know that $f \left(x\right)$ is increasing when $x = - 1$.