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Math Help - Arctan function

  1. #1
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    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!
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  2. #2
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    Quote Originally Posted by gralla55 View Post
    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]
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    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
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    Also, I thought it was f(-1) and f(0) that had to have the opposite signs, not the derivates...
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    Quote Originally Posted by gralla55 View Post
    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
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    Quote Originally Posted by gralla55 View Post
    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.
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  7. #7
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    Quote Originally Posted by gralla55 View Post
    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.
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    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.
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  9. #9
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    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)
    Last edited by skeeter; October 16th 2009 at 05:19 PM.
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    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!
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  11. #11
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    zeropoint ... a "root" or "zero"

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

    f(-1) and f(0) have opposite signs
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    Yes, the point I'm looking for is when f(x) = 0, not when f'(x) = 0. That should explain the confusion :P
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    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.
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