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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$\begingroup$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.
$\endgroup$ 2 $\begingroup$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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