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) = - (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!
$\displaystyle f(x) = -\frac{x}{2} + 3\arctan(1+2x) + \frac{1}{4}\ln(1+2x+2x^2)$
$\displaystyle f(-1) = \frac{1}{2} + 3\arctan(-1) + \frac{1}{4}\ln(1)$
$\displaystyle f(-1) = \frac{1}{2} - \frac{3\pi}{4}$
$\displaystyle f(0) = 3\arctan(1) = \frac{3\pi}{4}$
$\displaystyle 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]
If $\displaystyle f(-1)$ and $\displaystyle f(0)$ have opposite signs, then $\displaystyle f(x)=0$ for some $\displaystyle x\in(0,1)$. This is the intermediate value theorem.
Similarly, if $\displaystyle f'(-1)$ and $\displaystyle f'(0)$ have opposite signs, then $\displaystyle f'(x)=0$ for some $\displaystyle x\in(0,1)$, implying that $\displaystyle f$ has a local min/max in the interval, also known as a stationary point.
Right. I just did that and got the right answer, thank you!common denominator ... you have to do the grunt work algebra.
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.
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)
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!
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.