Question

The intervals of concavity and convexity of $f\left(x\right)=e^{-x^2}$ are

• A. Concave:$(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$
• B. Concave : $(-\infty ,-\frac{\sqrt{2}}{2})\cup (\frac{\sqrt{2}}{2},\infty )$
• C. Convex:$(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$
• D. Convex : $(-\infty ,-\frac{\sqrt{2}}{2})\cup (\frac{\sqrt{2}}{2},\infty )$

Answer 1

Answer: (A), (D)

Convex : $(-\infty ,-\frac{\sqrt{2}}{2})\cup (\frac{\sqrt{2}}{2},\infty )$

$f\left(x\right)=e^{-x^2}$
$f^{\prime }\left(x\right)=-2x{e}^{-x^2}$
$f^{\prime \prime }\left(x\right)=-2e^{-x^2}-2x{e}^{-x^2}(-2x)f^{\prime \prime }\left(x\right)=2e^{-x^2}(2x^2-1)$
$f^{\prime \prime }\left(x\right)>0\Rightarrow$ convex and $f^{\prime \prime }\left(x\right)<0\Rightarrow$ concave
So, convex in $x\in (-\infty ,-\frac{\sqrt{2}}{2})\cup (\frac{\sqrt{2}}{2},\infty )$
Concave in $x\in (-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$