# Is f(x) =e^x/x-x^3-3 concave or convex at x=-1?

Oct 1, 2016

$C o n v e x$

#### Explanation:

To check if the function is convex or concave we have to find$f ' ' \left(x\right)$

If $\textcolor{b r o w n}{f ' ' \left(x\right) > 0}$ then $\textcolor{b r o w n}{f \left(x\right)}$ is $\textcolor{b r o w n}{c o n v e x}$
If $\textcolor{b r o w n}{f ' ' \left(x\right) < 0}$ then $\textcolor{b r o w n}{f \left(x\right)}$ is $\textcolor{b r o w n}{c o n c a v e}$

first let us find $\textcolor{b l u e}{f ' \left(x\right)}$
$f ' \left(x\right) = \left(\frac{{e}^{x}}{x}\right) ' - \left({x}^{3}\right) ' - \left(3\right) '$
$f ' \left(x\right) = \frac{x {e}^{x} - {e}^{x}}{x} ^ 2 - 3 {x}^{2} - 0$
$\textcolor{b l u e}{f ' \left(x\right) = \frac{x {e}^{x} - {e}^{x}}{x} ^ 2 - 3 {x}^{2}}$

Now let us find $\textcolor{red}{f ' ' \left(x\right)}$
f''(x)= ((xe^x-e^x)'x^2-(x^2)'(xe^x-e^x))/(x^2)^2-6x
$f ' ' \left(x\right) = \frac{\left({e}^{x} + x {e}^{x} - {e}^{x}\right) {x}^{2} - 2 x \left(x {e}^{x} - {e}^{x}\right)}{x} ^ 4 - 6 x$
$f ' ' \left(x\right) = \frac{{x}^{3} {e}^{x} - 2 {x}^{2} {e}^{x} - 2 x {e}^{x}}{x} ^ 4 - 6 x$
Let us simplify the fraction by $x$

$\textcolor{red}{f ' ' \left(x\right) = \frac{{x}^{2} {e}^{x} - 2 x {e}^{x} - 2 {e}^{x}}{x} ^ 3 - 6 x}$

Now let us compute color(brown)(f''(-1)

$f ' ' \left(- 1\right) = \frac{{\left(- 1\right)}^{2} {e}^{- 1} - 2 \left(- 1\right) {e}^{- 1} - 2 {e}^{- 1}}{- 1} ^ 3 - 6 \left(- 1\right)$
$f ' ' \left(- 1\right) = \frac{{e}^{- 1} + 2 {e}^{- 1} - 2 {e}^{- 1}}{- 1} + 6$

$\textcolor{b r o w n}{f ' ' \left(- 1\right) = - {e}^{- 1} + 6}$
color(brown)(f''(-1)>0

So,$f ' ' \left(x\right) > 0$ at $x = - 1$

Therefore,$f \left(x\right)$ is covex at $x = - 1$

graph{e^x/x - x^3 -3 [-20, 20, -20, 20]}