# Calculus

Consider the function f(x)=xsqrt(36−x^2), −1≤x≤6.

This function has an absolute minimum value equal to what?

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3. 👁 52
1. f(x) = √(36-x^2) = (36-x^2)^(1/2)
f ' (x) = (1/2)(36-x^2)^(-1/2) (-2x)
= 0 for a max/min
thus x = 0 , f(0) = 6
but f(1) = √(36-1)
= √35 which is < 6

f(6) = √0 = 0

So the absolute minimum is 0

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posted by Reiny

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