# For what values of x is f(x)=-x^2+e^x concave or convex?

##### 1 Answer
Mar 25, 2018

$\left(- \infty , \ln 2\right) :$ Convex $\left(\ln 2 , \infty\right) : C o n c a v e$

#### Explanation:

Determining concavity requires finding $f ' ' \left(x\right)$ and examining where it is positive or negative.

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

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

To determine where this is positive or negative, let's set $f ' ' \left(x\right) = 0$ and solve.

$- 2 + {e}^{x} = 0$

${e}^{x} = 2$

$\ln \left({e}^{x}\right) = \ln \left(2\right)$

$x = \ln \left(2\right)$

Let's split up the domain of $f \left(x\right) , \left(- \infty , \infty\right) ,$ around $x = \ln \left(2\right) ,$ and test values for $f ' ' \left(x\right)$ to see if it is positive or negative:

$\left(- \infty , \ln \left(2\right)\right) : f ' ' \left(0\right) = - 2 + {e}^{0} = - 1 < 0$

Thus, since $f ' ' \left(x\right) < 0$ on $\left(- \infty , \ln \left(2\right)\right) , f \left(x\right)$ is convex on $\left(- \infty , \ln \left(2\right)\right)$.

$\left(\ln \left(2\right) , \infty\right) :$

$f ' ' \left(\ln \left(3\right)\right) = - 2 + {e}^{\ln} 3 = - 2 + 3 > 0$

Thus, since $f ' ' \left(x\right) > 0$ on $\left(\ln \left(2\right) , \infty\right) , f \left(x\right)$ is concave on $\left(\ln \left(2\right) , \infty\right)$.