Find the maximum and minimum values of f(x)=xe^(x^2/72) in the interval=(-2,12)

Expert Answers

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The function `f(x)=xe^(x^2/72)`

`f'(x) = e^(x^2/72) + x*((2x)/72)*e^(x^2/72)`

=> `e^(x^2/72)(1 + x^2/36)`

Solving f'(x) = 0 gives the equation `e^(x^2/72)(1 + x^2/36) = 0` which does not have any real roots.

There are no extreme points for `f(x) = x*e^(x^2/72)` lying in the given interval.

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