# 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

The minimum, maximum and inflection points will be at the points in which the derivative, in your case the integrand is equal to zero.

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 $x=-9$.

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@BabakSorouh - actually I'm having doubts. The integrand isn't defined at $x=\pi(n-1/2)$.. –  nbubis Jan 28 '13 at 7:43
So, those points would be regarded as critical points. –  B. S. Jan 28 '13 at 7:44
@BabakSorouh - the OP changed the question, and everything shold be defined properly. –  nbubis Jan 28 '13 at 7:49
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
@RezaM. - you need to find where the derivative, in this case the integrand is zero. the solution come from equating the integrand, and therefore the parabola to zero. Also fixed the solution according to 45 and not 25.. –  nbubis Jan 28 '13 at 7:51