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Math Help - Critical Numbers and Absolute Max/Min

  1. #1
    Member FalconPUNCH!'s Avatar
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    Critical Numbers and Absolute Max/Min

    1) Find the Critical Numbers for
    h(p) = \frac{p - 1}{p^2+4}

    h'(p) = \frac{p^2 + 4 - (p - 1)(2p)}{(p^2 + 4)^2}

    after all the work is done I get

    -p^2 + 2p + 4 = 0

    There aren't any critical numbers is what I got but I'm not sure.

    2) Find the Critical Numbers for  g(x) = \sqrt{1 - x^2}

    g'(x) = \frac{2x}{2 \sqrt{1-x^2}}

    \frac{x}{\sqrt{1-x^2}} = 0

    I get  x = 0 as a critical number but I'm not sure if it's right or the only answer

    3) Find the Absolute max and min for
     f(t) = 2cos(t) + sin(2t) in the interval [0 , \frac{pi}{2}]

    after simplifying the derivative for it I get:

     f'(t) = 2(cost(2t) + sin(t))

    I don't know what to do with it since I am bad with trig functions in these types of problems.
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  2. #2
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    Hello, FalconPUNCH!

    Here's #3 . . .


    3) Find the Absolute max and min for: . f(t) \:= \:2\cos t + \sin2t in the interval \left[0 ,\,\frac{\pi}{2}\right]

    The derivative is: . f'(t) \:=\:-2\sin t + 2\cos2t

    We have: . 2\cos2t - 2\sin t \;=\;0

    . . Then: . 2\left(1-2\sin^2t\right) - 2\sin t \;=\;0

    . . which simplifies to: . 2\sin^2\!t + \sin t = 1\;=\;0

    . . which factors: . (\sin t + 1)(2\sin t-1)\;=\;0


    Then we have: . \sin t + 1 \:=\:0\quad\Rightarrow\quad\sin t \:=\:-1\quad\Rightarrow\quad t \:=\:\frac{3\pi}{2} . . .
    not in the interval!

    And we have: . 2\sin t - 1\:=\:0\quad\Rightarrow\quad \sin t\:=\:\frac{1}{2}\quad\Rightarrow\quad t \:=\:\frac{\pi}{6}


    Find the value of the function at the critical value and the endpoints:

    f(0) \;=\;2\cos(0) + \sin(0) \;=\;2\!\cdot\!1+0\;=\;2

    f\left(\frac{\pi}{6}\right)\;=\;2\cos\left(\frac{\  pi}{6}\right) +\sin\left(\frac{\pi}{3}\right) \;=\;2\cdot\frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} \;=\;\frac{3\sqrt{3}}{2} \;\approx\;2.598 . ← absolute maximum

    f\left(\frac{\pi}{2}\right) \;=\;2\cos\left(\frac{\pi}{2}\right) + \sin(\pi) \;=\;2\!\cdot\!0 + 0 \;=\;0 . ← absolute minimum

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  3. #3
    Member FalconPUNCH!'s Avatar
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    Thank you. Can anyone check and make sure I didn't screw up on the first two?
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  4. #4
    Member FalconPUNCH!'s Avatar
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    Can anyone help? I have school in an hour and I won't be back the whole day.
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    1) Find the Critical Numbers for
    h(p) = \frac{p - 1}{p^2+4}

    h'(p) = \frac{p^2 + 4 - (p - 1)(2p)}{(p^2 + 4)^2}

    after all the work is done I get

    -p^2 + 2p + 4 = 0

    There aren't any critical numbers is what I got but I'm not sure.
    this quadratic equation has roots, two in fact, and therefore, there are two critical numbers.
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by FalconPUNCH! View Post
    2) Find the Critical Numbers for  g(x) = \sqrt{1 - x^2}

    g'(x) = \frac{2x}{2 \sqrt{1-x^2}}

    \frac{x}{\sqrt{1-x^2}} = 0

    I get  x = 0 as a critical number but I'm not sure if it's right or the only answer
    that is correct (this is the upper have of the circle with radius 1 centered at the origin. we have a horizontal slope at the top)
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  7. #7
    Member FalconPUNCH!'s Avatar
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    Quote Originally Posted by Jhevon View Post
    this quadratic equation has roots, two in fact, and therefore, there are two critical numbers.
    So do I just use the quadratic formula since I can't factor it?
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