Question

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Answer 1

to find an inflection point you need to take the second derivative of f(x) with respect to x and set it equal to zero. This is a requirement for an inflection point (but does not prove the point is an inflection point)

Answer 2

so \[\frac{df}{dx}=-2e^{-x^2}\]\[\frac{d^2f}{dx^2}=e^{-x^2}\left(4x^2-2\right)\]

Answer 3

isn"t [d^2f/dx^2=-2e^-x^2+4x^2e^x^2]

Answer 4

nope the chain rule gives. \[\frac{df}{dx}e^{-x^2}=\frac{d}{dx}\left(-x^2\right)e^{-x^2}=-2xe^{-x^2}\]

Answer 5

then you must use the product rule. so \[f''=-2x\frac{d}{dx}\left(e^{-x^2}\right)+\frac{d}{dx}\left(-2x\right)e^{-x^2}\]

Answer 6

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

Answer 7

\[f''=4x^2e^{-x^2}-2e^{-x^2}\] and then factoring out \(e^x\) you get \[f''=\left(4x^2-2\right)e^{-x^2}\]

Answer 8

now you should know that \(e^{-x^2}\) is never equal to zero. so just solve where \[4x^2-2=0\]\[x=\pm\frac{1}{\sqrt{2}}\]

Answer 9

ok now this particular function has something crazy about it

Answer 10

actually don't worry about it, but to check and see if they are inflection points, you must see if the second derivative changes signs at this point. which it does, so those are your two points.

Answer 11

do you know how to find the intervals where f(x)is concave downward

Answer 12

and the intervals where f(x)is concave upward

Answer 13

yes, you need to determine if \(f''(x)\) is positive or negative in that interval. so for the interval \(\left[-\frac{1}{2},\frac{1}{2}\right]\) you have a negative second derivative and the graph is concave down. the other two you have a positive second derivative and the graph is concave up.