# Is f(x)=(3-e^(2x))/x increasing or decreasing at x=-1?

Jan 10, 2017

$f$ decreases at $x = - 1$.

#### Explanation:

We know that, $f ' \left(a\right) < 0 \Rightarrow f \text{ is "darr" at } x = a$.

Now, $f \left(x\right) = \frac{3 - {e}^{2 x}}{x} = \left(3 - {e}^{2 x}\right) {x}^{-} 1$

To find $f ' \left(x\right)$, we use the Product and Chain Rule.

f'(x)={3-e^(2x)}d/dx(x^-1)+x^-1{d/dx(3-e^(2x)}

$= \left\{3 - {e}^{2 x}\right\} \left(- 1 {x}^{-} 2\right) + {x}^{-} 1 \left\{0 - {e}^{2 x} \frac{d}{\mathrm{dx}} \left(2 x\right)\right\}$

$= \frac{{e}^{2 x} - 3}{x} ^ 2 - \frac{2 {e}^{2 x}}{x}$

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

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

Since, $2 < e < 3 , 4 < {e}^{2} < 9 , \frac{1}{4} > \frac{1}{e} ^ 2 > \frac{1}{9} ,$ we have,

$f ' \left(- 1\right) < 0.$

Therefore, $f$ decreases at $x = - 1$.

Enjoy Maths.!