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

Oct 28, 2017

The function $f \left(x\right)$ is decreasing at $x = 2$

#### Explanation:

Calculate the first derivative and look at the sign

We need

$\left(\frac{u}{v}\right) ' = \frac{u ' v - u v '}{{v}^{2}}$

Here,

$u \left(x\right) = 1 - {e}^{x}$, $\implies$, $u ' \left(x\right) = - {e}^{x}$

$v \left(x\right) = {x}^{2}$, $\implies$, $v ' \left(x\right) = 2 x$

Therefore,

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

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

When $x = 2$

$f ' \left(2\right) = \frac{- 2 {e}^{2} - 2 + 2 {e}^{2}}{8} = - \frac{1}{4}$

As $f ' \left(2\right) < 0$, the function $f \left(x\right)$ is decreasing at $x = 2$

graph{(y-(1-e^x)/(x^2))(y-1000(x-2))=0 [-4.93, 12.85, -4.124, 4.765]}

Oct 28, 2017

$f \left(x\right) \text{ is decreasing at x =2}$

#### Explanation:

$\text{to determine if a function f(x) is increasing/decreasing at}$
$\text{x = a evaluate } f ' \left(a\right)$

• " if "f'(a)>0" then f(x) is increasing at x = a"

• " if "f'(a)<0" then f(x) is decreasing at x = a"

$\text{differentiate "f(x)" using the "color(blue)"quotient rule}$

$\text{given "f(x)=(g(x))/(h(x))" then}$

$f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2 \leftarrow \text{ quotient rule}$

$g \left(x\right) = 1 - {e}^{x} \Rightarrow g ' \left(x\right) = - {e}^{x}$

"h(x)=x^2rArrh'(x)=2x

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

$\Rightarrow f ' \left(2\right) = \frac{- 4 {e}^{2} - 4 + 4 {e}^{2}}{16} = - \frac{1}{4}$

$\text{since "f'(2)<0" then f(x) is decreasing at x = 2}$
graph{(1-e^x)/(x^2) [-10, 10, -5, 5]}