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

Mar 9, 2018

Function $f \left(x\right) = \frac{1 - x {e}^{x}}{1 - {x}^{2}}$ is increasing at $x = 2$

#### Explanation:

To find whether a function is increasing or decreasing at a given point say $x = {x}_{0}$, we need to differentiate it and find value of the derivative at $x = {x}_{0}$.

As $f \left(x\right) = \frac{1 - x {e}^{x}}{1 - {x}^{2}}$, using quotient formula

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

and at $x = 2$

$\frac{\mathrm{df}}{\mathrm{dx}} = \frac{- \left(1 - 4\right) \left({e}^{2} + 2 {e}^{2}\right) + 4 \left(1 - 2 {e}^{2}\right)}{1 - 4} ^ 2$

= $\frac{9 {e}^{2} + 4 - 8 {e}^{2}}{9} = \frac{{e}^{2} + 4}{9}$

As $\frac{\mathrm{df}}{\mathrm{dx}} > 0$, the function is increasing at $x = 2$.

graph{(1-xe^x)/(1-x^2) [-9.67, 10.33, -1.12, 8.88]}

Mar 9, 2018

Increasing

#### Explanation:

the sign of the derivative of the function determines if its increasing or decreasing

therefore, we have to differentiate it first,
I'll use the quotient rule here

therefore derivative =

$\frac{{e}^{x} \left({x}^{3} - {x}^{2} - x - 1\right) + 2 x}{1 - {x}^{2}} ^ 2$

therefore, just put in x = 2

$\frac{{e}^{2} \left(8 - 4 - 2 - 1\right) + 2 \cdot 2}{9}$

this value is positive, therefore
the function is increasing at $x = 2$