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Question 666176: Consider the function:
y(x)= e(-x^2)
Calculate and draw a sign diagram for the first derivative. Where is the function
increasing or decreasing. Are there any peaks or troughs? Does the function have an
(absolute) maximum.
ii) Calculate and draw a sign diagram for the second derivative. Where is the function
convex or concave. Are there any ináection points?
Answer by swincher4391(1107)   (Show Source):
You can put this solution on YOUR website! y'(x) = -2x e^(-x^2)
Your critical point is where 0 = -2xe^(-x^2)
Since e will never be 0, we have to rely on -2x, which of course, 0 makes it 0.
So <------------0---------------> If we plug in a negative number we get a positive, if we plug in a positive we get a negative so we get:
<---negative----0------positive---->
From -oo to 0 decreasing, trough at 0, increasing to infinity. No absolute max, since we have a trough. A peak would give us an abs max.
y''(x) = -2x*-2x*e^(-x^2) + e^(-x^2)*-2 (by product rule)
=4x^2e^(-x^2) - 2e^(-x^2)
= (4x^2-1)(e^(-x^2))
4x^2 -1 = 0
4x^2 = 1
x^2 = 1/4
x = +- 1/2
<---- -1/2 ------------ 1/2 ---------->
choose -1 for the left side.
We get positive.
choose 0, we get negative
choose 1 we get positive
<--positive -- -1/2 ---- negative ---- 1/2 ---- positive---->
Hence we have concave up from negative infinity to -1/2
concave down from -1/2 to 1/2
concave up from 1/2 to oo
points of inflection are x= -1/2 and 1/2
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