1. ## integral..find local max

$\displaystyle f(x) = \int_0^x \frac{ t^2 - 36 }{ 1+\cos^2(t)} dt$

At what value of $\displaystyle x$ does the local max of $\displaystyle f(x)$ occur?

$\displaystyle x = ?$

$\displaystyle f '(x) = \int_0^x \frac{ x^2 - 36 }{ 1+\cos^2(x)}$
im stuck on this part, not sure what to do after.

2. Originally Posted by viet
$\displaystyle f(x) = \int_0^x \frac{ t^2 - 36 }{ 1+\cos^2(t)} dt$

At what value of $\displaystyle x$ does the local max of $\displaystyle f(x)$ occur?
If a real function is differenciable then the local max (relative maximum) must occurs when $\displaystyle f'(x)=0$

Using the Fundamental theorem,
$\displaystyle f'(x)=\frac{x^2-36}{1+\cos^2 x}$
Thus,
$\displaystyle \frac{x^2-36}{1+\cos^2 x}=0$
Thus,
$\displaystyle x^2-36=0, x=\pm 6$
Those are the possible points.
To determine what status they have we will use the first-derivative test.

Consider the three intervals,
$\displaystyle x<-6$----> $\displaystyle f'(x)<0$----> decreasing.
$\displaystyle -6<x<6$----> $\displaystyle f'(x)>0$----> increasing.
$\displaystyle 6<x$----> $\displaystyle f'(x)<0$----> decreasing.

Thus, the relative maximum occurs at $\displaystyle x=6$

3. Originally Posted by viet
$\displaystyle f(x) = \int_0^x \frac{ t^2 - 36 }{ 1+\cos^2(t)} dt$

At what value of $\displaystyle x$ does the local max of $\displaystyle f(x)$ occur?

$\displaystyle x = ?$

$\displaystyle f '(x) = \int_0^x \frac{ x^2 - 36 }{ 1+\cos^2(x)}$
im stuck on this part, not sure what to do after.
If:

$\displaystyle f(x) = \int_0^x \frac{ t^2 - 36 }{ 1+\cos^2(t)} dt$

then (fundamental theorem of calculus)

$\displaystyle f'(x) = \frac{ x^2 - 36 }{ 1+\cos^2(x)}$

since $\displaystyle { 1+\cos^2(x)} \ne 0$ the stationary points of $\displaystyle f(x)$ are the roots of $\displaystyle x^2-36=0$, which are $\displaystyle x=\pm 6$.

To determine which is the maximum and which the minimum find $\displaystyle f''(x)$ at $\displaystyle x=\pm 6$, and check the sign negative and its a maximum and positive its a minimum (the minimum is at $\displaystyle -6$ but you had better check)

RonL