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Math Help - Finding critical number for derivative of exponential?

  1. #1
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    Finding critical number for derivative of exponential?

    Ok, so I have a function that was originally f (x) = e^2x - 3e^x + 2....the problem asks me to find X intercepts, Y intercepts, relative extrema, inflection points, concavity and any horizontal asymptotes...and if you're wondering why all that, it's so I can finally graph it.

    So I factored it and determined that it's (e^x - 1) * (e^x - 2), so e^x = 1 and e^x = 2, which means x = ln1 and x = ln2, so there's X-intercepts at (0,0) and (0.693,0). So I got that ok.

    Next for the Y-int, I set x equal to 0 and solved for it, and determined e^2(0) - 3e^(0) + 2 = 0 so Y-intercept is also at (0,0).

    Now for the relative extrema, this is where I got stuck. I determined the derivative to be f ' (x) = 2e^2x - 3e^x.....and this is where I'm stuck. What would be the best way to factor this out so I can get the critical numbers of the first derivative? If I factor out a e^x, then I have e^x ( 2e^x - 3) which doesn't necessarily help me since if you set the stuff inside parentheses to x=0 then you have ln3/2 but the e^x on the outside, well even if e^(0), it still equals 1...so yeah, I'm probably overlooking something simple, but what are the critical numbers for the first derivative? And more importantly, how did you find them? Thanks guys!
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  2. #2
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    The derivative of said function is

    2e^{2x}-3e^{x}

    2(e^{x})^{2}-3e^{x}=0

    Let u=e^x:

    2u^{2}-3u=0

    u(2u-3)=0

    u=0 \;\ or \;\ \frac{3}{2}

    e^{x}=0, undefined.

    e^{x}=\frac{3}{2}, \;\ x=ln(3/2)
    Last edited by galactus; November 23rd 2007 at 05:32 PM.
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  3. #3
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    Quote Originally Posted by galactus View Post
    The derivative of said function is

    2e^{2x}-3e^{x}

    2(e^{x})^{2}-3e^{x}=0

    Let u=e^x:

    2u^{2}-3u=0

    u(2u-3)=0

    u=0 \;\ or \;\ \frac{3}{2}

    e^{x}=0, \;\ x=1

    e^{x}=\frac{3}{2}, \;\ x=ln(3/2)
    Ok, so the critical numbers are x=ln(3/2) as well as x=1? Thanks...so then for the second derivative, f " (x) = 4e^{2x}-3e^{x} so factoring it, I have e^{x}(4e^{x}-3) so the critical numbers of the second derivative for concavity would be x=1 and x=ln(3/4), correct?

    Also, am I correct in assuming that there is a horizontal asymptote at y=2 due to the fact that a normal exponential function has a H.A. at y=0 but the original function has a + 2 constant on the end, thus moving everything (including the H.A.) up by two?
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  4. #4
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    I made a typo. e^1 does not equal 0. I fix. e^x never equals 0.

    ln(3/4) is an inflection point, but 1 is not. There is a minimum at x=ln(3/2), but not at x=1.

    Also, \lim_{x\rightarrow{-\infty}}f(x)=2

    Yes, there is a HA at y=2.
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  5. #5
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    Quote Originally Posted by galactus View Post
    I made a typo. e^1 does not equal 0. I fix. e^x never equals 0.

    ln(3/4) is an inflection point, but 1 is not. There is a minimum at x=ln(3/2), but not at x=1.

    Also, \lim_{x\rightarrow{-\infty}}f(x)=2

    Yes, there is a HA at y=2.
    Ok, that makes sense about the e^x thing...thank you so much!
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