f(x) = e^-x^2
f'(x) = e^-x^2 + (-2x)
f'(x) = -2xe^-x^2
f"(x) = -2x3^-x^2 +e^-x^2 + (-2x)
f"(x) =-4x(2e^-x^2)
how do you solve these types (e^-x^2) of problems to find the inflection point and concavity?
Hello,
I'm not quite certain if I read your problem correctly...
$\displaystyle f(x)=e^{-x^2}$ . Use chain rule to get the first derivation:
$\displaystyle f'(x) = e^{-x^2} \cdot (-2x) = -2x\cdot e^{-x^2}$
f'(x) = 0 if x = 0
Use product rule to get the second derivation:
$\displaystyle f''(x) = e^{-x^2} \cdot (-2) + (-2x) \cdot e^{-x^2} \cdot (-2x) = e^{-x^2} ( 4x^2 -2)$
f''(x) = 0 if 4x² - 2 = 0 that means $\displaystyle x=\sqrt{\frac{1}{2}} \ \vee \ x= -\sqrt{\frac{1}{2}}$