Suppose

Question

f(x)=∫ (from 0 to x) (t^2-36)/2+cos^2(t) dt.

For what value(s) of x
does f(x) have a local maximum?

Answer 1

well, df/dx = (x^2-36)/2 + cos^2(x)

where is that zero?
which is a max?

Answer 2

How do we find the zero is it with the denominator or the numerator ?

Answer 3

well, there's no algebraic way to do it that I know of. Best to check the graph and approximate it. You know that x^2-36 is zero at ±6. cos^2(x) is positive there, so x will be on the inside of the interval (-6,6). So you can check nearby values and approximate it as closely as you want.

Now you have to decide which is a max and which is a min when f'=0. You know that at a maximum, the slope changes from positive to negative (top of a hill). So, since f'(-6) > 0 and f'(-5) < 0, the value near -6 is a maximum. The graph of f' is at

http://www.wolframalpha.com/input/?i=(x%5E2-36)%2F2+%2B+cos%5E2(x)