Find all critical numbers and use 2nd derivative test to determine all local extrema

Question

anonymous · Answer

$$f(x)=e ^{-x ^{2}}$$

anonymous · Answer

For the first derivative i got  $$f'(x)=-2e ^{-x ^{2}}x$$

anonymous · Answer

where x must equal 0 for the function to equal 0

anonymous · Answer

second derivative i got $$e ^{-x ^{2}}(4x ^{2}-2)$$

anonymous · Answer

put in 0 and i get -2. So what does this tell me? lol

anonymous · Answer

is -2 a local max or min?

anonymous · Answer

After finding the first derivative of $$f'(x) = -2xe ^{-x ^{2}}$$,
plug in 0 to find all critical points. You get x-values of -0.707, 0, and 0.707. This means those are your three critical points. Then, find the second derivative. $$f''(x) = -2e ^{-x ^{2}}+(-2e ^{-x ^{2}})^{2}$$. Set that equal to zero for the second derivative test, and get at -.707, y = 0, at 0, y = -2, and at .707, y = 0. So, x = 0 is a maximum, x = +/- .707 are points of inflection :)

anonymous · Answer

i dont see how you got =/-.707 when setting the first derivative equal to 0

anonymous · Answer

+\-**

anonymous · Answer

$$f′(x)=−2xe ^{-x ^{2}}$$
Sorry, x = 0 is the only value that satisfies y = 0, so, x = 0 is a maximum. x = +/- .707 is where the SECOND derivative is 0, making the POI's. Sorry, I should have clarified.