# Local Max/Min, Critical points of integral

Given $$f(t) = \int_0^t \frac{x^2+14x+45}{1+\cos^2(x)}dx$$

I need to find the local max of f(t). Well here using the fundamental theorem of calculus, I know I can just replace the $x$ with $t$. But I do not remember how to find the local max/min and if I remember correctly critical points were in the same context, So some insight on critical points would be good too. Thank you.

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Hint: What is $f'(t)$? When is it zero? –  Alexander Thumm Jan 28 '13 at 7:33

Just find $f''(t)$ and then see the sign of $f''(t)$ at the critical points. You should get $f''(-5)>0$ which tells you $x=-5$ is a minima and $f''(-9)<0$ which tells you $x=-9$ is a maxima. See second derivative test.

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I occasionally, use this test in the class. +1 –  B. S. Jan 28 '13 at 8:13
@BabakSorouh: Thank you. –  Mhenni Benghorbal Jan 28 '13 at 8:26
In your case, these are simply the solutions to: $$x^2+14x+45=0$$ Namely, $x=-9, -5$. To find which is a minimum / maximum, I would just evaluate the integrand at some sample points such as $x=0,-2\pi,-3\pi$. You get that for instance: $$f'(0) = \frac{45}{2} >0$$ And that: $$f'(-2\pi) = \frac{4\pi^2-28\pi+45}{2} <0$$ This means the point $x=-5$ is a minimum, since the derivative is increasing at between $-2\pi$ and $0$. A similar calculation follows for the point $x=-9$.
@BabakSorouh - actually I'm having doubts. The integrand isn't defined at $x=\pi(n-1/2)$.. –  nbubis Jan 28 '13 at 7:43
Confused: I need to find where my solution = 0 ? where did $x = -7 +-2sqrt(6)$ come from ? –  Reza M. Jan 28 '13 at 7:50