# If the function #f# is defined by #f(x)=int_0^x-sint^2dt# on the closed interval #-1<=x<=3#, then #f# has a local maximum at x=?

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The correct answer is 2.507. But how?

The correct answer is 2.507. But how?

##### 1 Answer

Apr 19, 2018

Taking the derivative, we get

#f'(x) = -sin(x^2)#

We must now find the critical numbers.

#0 = -sin(x^2)#

#0, pi, 2pi, ... = x^2#

#x = 0, sqrt(pi), sqrt(2pi)#

Since the derivative at

It follows that

Hopefully this helps!