No idea of what to do here...

**ckoeber** · January 29th 2010, 06:55 AM

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.

**Plato** · January 29th 2010, 07:03 AM

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

**General** · January 29th 2010, 07:04 AM

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.

**ckoeber** · January 29th 2010, 07:08 AM

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?

**ckoeber** · January 29th 2010, 07:09 AM

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?

**Plato** · January 29th 2010, 07:34 AM

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

**drumist** · January 29th 2010, 08:46 AM

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.

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

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

$\implies x^2-1 = 0$

**ckoeber** · January 29th 2010, 08:49 AM

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.

**ckoeber** · January 29th 2010, 08:50 AM

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.

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