# Thread: No idea of what to do here...

1. ## No idea of what to do here...

Hello,

I have no idea on how to solve the problem below. Do I take the intergral or do I set the given derivative equal to something?

Thanks!

Regards,
Chris K.

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

3. Originally Posted by ckoeber
Hello,

I have no idea on how to solve the problem below. Do I take the intergral or do I set the given derivative equal to something?

Thanks!

Regards,
Chris K.
See this formula:
If $f(x)=\int_a^x g(t)dt$
then $f'(x)=g(x)$
By applying this formula, you will get the derivative.
Now, It should be easy.

4. Originally Posted by Plato
Hint: $f'(x)=\frac{x^2-1}{1+\cos^2(x)}$
I figured that, but then I go ahead and set that to zero, right?

I graphed the function and it is periodic so I am assuming that I need to find the answer in relations to pi.

Am I correct?

5. Originally Posted by General
See this formula:
If $f(x)=\int_a^x g(t)dt$
then $f'(x)=g(x)$
By applying this formula, you will get the derivative.
Now, It should be easy.
Thanks. Essentially I substitute the limit into the orginal function and work it out. But in this problem, as I mentioned to Plato, the graph is periodic.

So the answer would be in relations to pi somehow, right?

6. $\begin{gathered}
f'( - 1) = 0,\;f'( - 1^ - ) > 0,\;f'( - 1^ + ) < 0 \Rightarrow \quad \max \hfill \\
f'(1) = 0,\;f'(1^ - ) < 0,\;f'(1^ + ) > 0 \Rightarrow \quad \min \hfill \\
\end{gathered}$

7. Originally Posted by ckoeber
So the answer would be in relations to pi somehow, right?
No, because we just need to find where it equals zero.

$f'(x)=\frac{x^2-1}{1+\cos^2(x)}=0$

$\implies \frac{x^2-1}{1+\cos^2(x)} \cdot (1+\cos^2(x)) = 0 \cdot (1+\cos^2(x))$

$\implies x^2-1 = 0$

8. Originally Posted by drumist
No, because we just need to find where it equals zero.

$f'(x)=\frac{x^2-1}{1+\cos^2(x)}=0$

$\implies \frac{x^2-1}{1+\cos^2(x)} \cdot (1+\cos^2(x)) = 0 \cdot (1+\cos^2(x))$

$\implies x^2-1 = 0$
Thanks! This is what I was looking for! I thought I couldn't solve for x because the denominator was throwing me off.

I feel dumb as all it took was simple arithmetic to solve

Regards,
Chris K.

9. Originally Posted by Plato
$\begin{gathered}
f'( - 1) = 0,\;f'( - 1^ - ) > 0,\;f'( - 1^ + ) < 0 \Rightarrow \quad \max \hfill \\
f'(1) = 0,\;f'(1^ - ) < 0,\;f'(1^ + ) > 0 \Rightarrow \quad \min \hfill \\
\end{gathered}$
After another poster showed me how to solve I see how to came to this conclusion now.

Thanks.