Question

Find the intervals of convexity and concavity of the Gaussian curve $y=e^{-x^2}$ and also find the point of inflection

Answer 1

$y=e^{{-x}^2}$

$y^{\prime }=-2x{e}^{{-x}^2}$

$y^{\prime \prime }=e^{{-x}^2}(4x^2-2)$

$y^{\prime \prime }=0,x=\pm \frac{1}{\sqrt{2}}$

$y^{\prime \prime }>0forx<\frac{-1}{\sqrt{2}}andx>\frac{1}{\sqrt{2}}$

$y^{\prime \prime }<0forx\in (\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

So it is concave up in $x\in (-\infty ,\frac{-1}{\sqrt{2}})\cup (\frac{1}{\sqrt{2}},\infty )$

Concave downwards in $x\in (\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}})$

And $x=\frac{-1}{\sqrt{2}},\frac{1}{\sqrt{2}}$ are inflection points.

So $(\frac{-1}{\sqrt{2}},e^{\frac{-1}{2}}),(\frac{1}{\sqrt{2}},e^{\frac{-1}{2}})$ are inflection points.