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Math Help - Apps of Differentiation help

  1. #1
    Member FalconPUNCH!'s Avatar
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    Unhappy Apps of Differentiation help

    Need help finding (a) Critical Numbers and (b) Inflection points for these two

    1.  y = 2x - tanx ;  \frac{-pi}{2} < x < \frac{pi}{2}

     y' = 2 - sec^2(x)

     y'' = -sec(x)tan(x)

    2.  y = e^(-x) sinx ; 0 < or equal to x < or equal to 2pi

     y' = -e^x cosx

     y'' = -e^x(-sinx) <--- Not sure if that's correct

    I'm not very fond of trig functions and I need help with these two. I need help asap
    Last edited by FalconPUNCH!; November 29th 2007 at 05:26 PM.
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  2. #2
    Member FalconPUNCH!'s Avatar
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    I also need help with when they concave down and up. I need help as soon as possible.
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  3. #3
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    Need help finding (a) Critical Numbers and (b) Inflection points for these two
    crtical numbers are the solutions of f'(x) = 0

    the possible inflection points are the solution of f''(x) = 0
    to check whether it is an inflection point, try to check the following:
    let x_o be a possible point of inflection.
    if f''(x< x_o) and f''(x > x_o) are of different signs, then (x_o, f(x_o)) is an inflection point.
    for concavity, if x_o is a possible point of inflection, and if f''(x > x_o) > 0, then the fucntion is concave up on the interval x > x_o; if f''(x > x_o) < 0, then the fucntion is concave down on the interval x > x_o

    Quote Originally Posted by FalconPUNCH! View Post

    1.  y = 2x - tanx ;  \frac{-pi}{2} < x < \frac{pi}{2}

     y' = 2 - sec^2(x)

     y'' = -sec(x)tan(x)
    so,  2 - \sec ^2 x = 0 \implies 2 = \sec ^2 x \implies \sqrt 2 = \sec x \implies \cos x = \frac{\sqrt 2}{2} \implies x = \frac{\pi}{4} ..
    and
     y'' = -2\sec ^2 x \, \tan x = 0 \implies \sec ^2 x = 0 or  \tan x = 0, \implies x = 0 (sec x is never zero)
    Quote Originally Posted by FalconPUNCH! View Post

    2.  y = e^(-x) sinx ; 0 < or equal to x < or equal to 2pi

     y' = -e^x cosx

     y'' = -e^x(-sinx) <--- Not sure if that's correct

    I'm not very fond of trig functions and I need help with these two. I need help asap
    i think, your function is supposed to be  y = e^{-x} \sin x
    so,  y' = -e^{-x} \sin x + e^{-x} \cos x and
    y'' = - (-e^{-x} \sin x + e^{-x} \cos x) + (-e^{-x} \cos x - e^{-x} \sin x) = -2e^{-x} \cos x
    Last edited by kalagota; November 29th 2007 at 06:24 PM.
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  4. #4
    Member FalconPUNCH!'s Avatar
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    Well I know how to find them but with Trig Functions it's harder for me to do. Am I supposed to use radians? I'm not too sure.
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  5. #5
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    Well I know how to find them but with Trig Functions it's harder for me to do. Am I supposed to use radians? I'm not too sure.
    i've edited it.. did you see it?
    hmm, yes, you have to do the angles in radians..
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  6. #6
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by kalagota View Post
    i've edited it.. did you see it?
    hmm, yes, you have to do the angles in radians..
    Oh I was unsure. Thanks for your help
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  7. #7
    Member FalconPUNCH!'s Avatar
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    I'm having problems with the local max for the 2nd problem. I think there's another critical number I'm missing but I got \frac{pi}{2} * as a critical number. I'm not too sure if there's supposed to be another.


    *It's probably wrong as well.
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  8. #8
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    I'm having problems with the local max for the 2nd problem. I think there's another critical number I'm missing but I got \frac{pi}{2} * as a critical number. I'm not too sure if there's supposed to be another.


    *It's probably wrong as well.
    can you show us the solution..
    i think, it was at max if  x = \frac{\pi}{4}
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  9. #9
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by kalagota View Post
    can you show us the solution..
    I would if I wrote it down. I'm doing it with a calculator right now but I'm going to try again. My trig teacher wasn't that great (was new at teaching) so I don't understand it fully because of the way he taught.
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  10. #10
    Member FalconPUNCH!'s Avatar
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    Well I don't know what to do so I give up. I've looked in the book for help and asked someone else but no luck. Thanks for helping anyways.

    I did get that \frac{pi}{2} is a critical number though.

    sin(x) = 0 ==> x = \frac{pi}{2} not sure if it's correct though
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  11. #11
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    Well I don't know what to do so I give up. I've looked in the book for help and asked someone else but no luck. Thanks for helping anyways.

    I did get that \frac{pi}{2} is a critical number though.

    sin(x) = 0 ==> x = \frac{pi}{2} not sure if it's correct though
    to get the critical number, you have to equate the first derivative to 0, not the function itself..
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  12. #12
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    2.  y = e^(-x) sinx ; 0 < or equal to x < or equal to 2pi
    y^{\prime} = -e^{-x}sin(x) + e^{-x}cos(x)

    For critical numbers we have
    -e^{-x}sin(x) + e^{-x}cos(x) = 0

    -sin(x) + cos(x) = 0

    sin(x) = cos(x)

    tan(x) = 1

    This occurs when x = \frac{\pi}{4} and x = \frac{5 \pi}{4}.

    y^{\prime \prime} = e^{-x}sin(x) - e^{-x}cos(x) - e^{-x}cos(x) + e^{-x} \cdot -sin(x)

    y^{\prime \prime} = e^{-x}sin(x) - 2e^{-x}cos(x) - e^{-x}sin(x)

    y^{\prime \prime} = -2e^{-x}cos(x)

    Inflection points are where y^{\prime \prime} = 0, so
    -2e^{-x}cos(x) = 0

    cos(x) = 0

    This occurs when x = \frac{\pi}{2} and x = \frac{3 \pi}{2}

    -Dan
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  13. #13
    Member FalconPUNCH!'s Avatar
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    I'm so stupid lol! I didn't set sinx = cosx and didn't know that I could do that. >_< I really need to go back and look over trig functions. Also I didn't even notice I was using the original function and not the derivative till you told me kalagota. Alright I'm gonna go back and finish the rest.

    Tops, I don't add 360 to \frac{pi}{4} because the interval is 0 < or equal to x < or equal to 2pi?
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  14. #14
    Member FalconPUNCH!'s Avatar
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    Is it possible if one of you can give me a site with trig function identities and rules? I don't want to be a bother next time I come around them.
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