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Thread: Critical points

  1. #1
    Junior Member
    Jan 2008

    Critical points

    Ok I've worked on these for a while and can't seem to come up with the right answer. Can someone help me out?

    1. f(x) = x^2/1-x
    Find the critical points

    2. f(x) = x^6 - 3x^4 + 3x^2 - 1
    Find the critical points

    Were also asked to find the domain, intercepts, symmetry and asymptotes. I'm pretty sure there aren't asymptotes for either one, and I'm pretty sure I got the answers to the others. If you want to answer those to to verify my answers that would be great. Thanks!
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  2. #2
    Global Moderator

    Nov 2005
    New York City
    I can see your first mistake , "there are no asymptotes", there is an asymptote for $\displaystyle f(x)=\frac{x^2}{1-x}$ at $\displaystyle x=1$ with $\displaystyle \lim_{x\to 1^-} f(x) = +\infty$ and $\displaystyle \lim_{x\to 1^+}f(x) = -\infty$. However, there are no horizontal asymptotes because $\displaystyle \lim_{x\to \infty}f(x) =\lim_{x\to -\infty}f(x) = +\infty$.
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  3. #3
    Senior Member
    Dec 2007
    To find the critical points, take the derivative of the function and set it equal to 0. Then solve for x.

    1. f(x) = x^2/1-x
    f'(x) = \frac {2x(1-x)-(-1)x^2}{(1-x)^2} = 0$
    \frac {2x-2x^2+x^2}{(x-1)^2} = 0$
    $\displaystyle 2x-x^2 = 0, x \not = 1$

    So the critical points are (2,f(2)) and (0, f(0))
    So the final answer is (2,-4) and (0, 0)

    I will leave you to do the second question
    Last edited by badgerigar; Jan 20th 2008 at 08:46 PM. Reason: fixed latex error
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