1. ## Arctan function

f(x) = - (x/2) + 3arctan(1+2x) + (1/4)ln(1+2x+2x^2)

What is f(-1) and f(0), and how do I show that f has a stationary point in the interval (-1,0)? What is f'(x)?

It's the arctan(1+2x) that really confuses me. I'd be very grateful for any help!

2. Originally Posted by gralla55
f(x) = - (x/2) + 3arctan(1+2x) + (1/4)ln(1+2x+2x^2)

What is f(-1) and f(0), and how do I show that f has a stationary point in the interval (-1,0)? What is f'(x)?

It's the arctan(1+2x) that really confuses me. I'd be very grateful for any help!
$f(x) = -\frac{x}{2} + 3\arctan(1+2x) + \frac{1}{4}\ln(1+2x+2x^2)$

$f(-1) = \frac{1}{2} + 3\arctan(-1) + \frac{1}{4}\ln(1)$

$f(-1) = \frac{1}{2} - \frac{3\pi}{4}$

$f(0) = 3\arctan(1) = \frac{3\pi}{4}$

$f'(x) = -\frac{1}{2} + \frac{6}{1 + (1+2x)^2} + \frac{1}{4} \cdot \frac{2+4x}{1+2x+2x^2}$

if f'(-1) and f'(0) have opposite signs, f'(0) = 0 in the interval [-1.0]

3. Thank you very much! I do know f'(x) is supposed to be (3-x^2)/(1+2x+2x^2) though, just not sure how to get there :P

4. Also, I thought it was f(-1) and f(0) that had to have the opposite signs, not the derivates...

5. Originally Posted by gralla55
Also, I thought it was f(-1) and f(0) that had to have the opposite signs, not the derivates...
a stationary point occurs when f'(x) = 0

6. Originally Posted by gralla55
Also, I thought it was f(-1) and f(0) that had to have the opposite signs, not the derivates...
If $f(-1)$ and $f(0)$ have opposite signs, then $f(x)=0$ for some $x\in(0,1)$. This is the intermediate value theorem.

Similarly, if $f'(-1)$ and $f'(0)$ have opposite signs, then $f'(x)=0$ for some $x\in(0,1)$, implying that $f$ has a local min/max in the interval, also known as a stationary point.

7. Originally Posted by gralla55
Thank you very much! I do know f'(x) is supposed to be (3-x^2)/(1+2x+2x^2) though, just not sure how to get there :P
common denominator ... you have to do the grunt work algebra.

8. common denominator ... you have to do the grunt work algebra.
Right. I just did that and got the right answer, thank you!

Still, I'm not quite getting the intermediate value theorem. But I'll try:

f'(-1)= (3-((-1)^2)) / (1+(2*-1)+(2*(-1)^2))) = 2

f'(0)= (3-((0)^2)) / (1+(2*0)+(2*(0)^2))) = 3

Which should tell me something I think.

9. then f(x) might not have a stationary point on the interval (-1,0) ...

as a matter of fact, after graphing f(x), I can see that it doesn't.

are all problem statements correct?

... looks like an inflection point in (-1,0)

10. On second thought I think "stationary point" is the wrong translation for what I'm looking for.

Let's see... the text says:

"Find f(-1) and f(0), and use the intermediate value theorem to show that f has a ...point in the interval (-1,0)"

It literally translates zeropoint but that's not a word I think. Perhaps inflection point is the right word. Lol, I must be superstressed. Thanks again for helping me out!

11. zeropoint ... a "root" or "zero"

f(x) does have a root in (-1,0)

f(-1) and f(0) have opposite signs

12. Yes, the point I'm looking for is when f(x) = 0, not when f'(x) = 0. That should explain the confusion :P

13. So I'm starting to get it now. The last part of the problem is to explain where the function is growing and where it is shrinking. That would be where f'(x) = 0 I guess. Which means:

f'(x) = 0
(3-x^2)/(1+2x+2x^2) = 0
3/(1+2x+2x^2) = x^2/(1+2x+2x^2)
3 = x^2
x = +-(root)3

And then I'm supposed to explain, without any calculations, if f has any zero-points when x>0, and if so, how many. This is where I'm a little unsure again.