# Is f(x)=-sqrt(e^x -x^3-2x  concave or convex at x=-1 ?

May 15, 2017

$f \left(x\right)$ is convex at $x = - 1$.

#### Explanation:

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

Determine the concavity of a graph by finding the second derivative of the function, and the sign of f"(-1) represents the concavity.

Differentiate using the chain rule:
$f ' \left(x\right) = - \frac{1}{2} {\left({e}^{x} - {x}^{3} - 2 x\right)}^{- \frac{1}{2}} \left({e}^{x} - 3 {x}^{2} - 2\right)$

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

Differentiate using the quotient rule:
$f ' ' \left(x\right) = \frac{- 2 \left({e}^{x} - {x}^{3} - 2 x\right) \left({e}^{x} - 6 x\right) + {\left({e}^{x} - 3 {x}^{2} - 2\right)}^{2} \left(2\right)}{4 {\left({e}^{x} - {x}^{3} - 2 x\right)}^{2}}$

Plug in $x = - 1$:
$f ' ' \left(- 1\right) = \frac{- 2 \left(\frac{1}{e} + 1 + 2\right) \left(\frac{1}{e} + 6\right) + {\left(\frac{1}{e} - 3 - 2\right)}^{2} \left(2\right)}{4 {\left(\frac{1}{e} + 3\right)}^{2}}$

This value is negative because the numerator is negative while the denominator is positive.

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