1. ## integral..find local max

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

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

$x = ?$

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

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

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

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

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

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

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

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

$x = ?$

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

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

then (fundamental theorem of calculus)

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

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

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

RonL