`f(x) = x e^(-x^2/8), [-1, 4]` Find the absolute maximum and minimum values of f on the given interval

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Chapter 4, 4.1 - Problem 59 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsarora17 | (Level 2) Associate Educator

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`f(x)=xe^(-x^2/8)`

differentiating by applying product rule,

`f'(x)=x(e^(-x^2/8))((-2x)/8)+e^(-x^2/8)`

`f'(x)=e^(-x^2/8)(-x^2/4+1)`

`f'(x)=-1/4e^(-x^2/8)(x^2-4)`

Now to find the absolute extrema of the function , that is continuous on a closed interval, we have to find the critical numbers that are in the interval and evaluate the function at the endpoints and at the critical numbers.

Now to find the critical numbers, solve for x for f'(x)=0.

`-1/4e^(-x^2/8)(x^2-4)=0`

`x^2-4=0 , x=+-2`

`e^(-x^2/8)=0 , No solution`

The critical number -2 is not in the interval (-1,4)

So evaluating the function at the critical number 2 and at the end points of the interval,

`f(-1)=-1e^(-1/8)=-1/e^(1/8)`

`f(4)=4e^(-4^2/8)=4e^-2=4/e^2`

`f(2)=2e^(-2^2/8)=2e^(-1/2)=2/sqrt(e)`

So the function has absolute maximum=2/`sqrt(e)` at x=2

Function has no absolute minimum , see graph.

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gsarora17 | (Level 2) Associate Educator

Posted on

Pl. note correction in absolute minimum.

Function has absolute minimum= `-1/root(8)(e)`     at x=-1

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