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Thread: critical points

  1. #1
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    critical points

    Hi;
    Do all critical points ie local max/local min have a slope of zero.

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  2. #2
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    Re: critical points

    No. This is wrong for two reasons. First a "critical point" is defined in any Calculus text as a point at which the derivative is 0 or does not exist. For example, the absolute value function, y= |x|, has derivative equal to 1 for x> 0, -1 for x< 0, never 0. x= 0 is, however, a critical point because the derivative does not exist there. Second your "ie local max/local min" is wrong because a critical point is not necessarily either a local maximum or a local minimum. $\displaystyle y= x^3$ has derivative $\displaystyle y'= 3x^2$ which is 0 at x= 0 so x= 0 is a critical point. But $\displaystyle y= x^3$ has neither a local maximum not a local minimum there.
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  3. #3
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    Re: critical points

    Ok so why isnt my tangent zero (tis is my first look at derivatives and calculus) so need some help.

    I posted my graph to help.


    critical points-tangent.jpg



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  4. #4
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    Re: critical points

    Looks like $f(x)=x^3-4x^2-5x+36$

    So, $f'(x)=3x^2-8x-5$

    $3x^2-8x-5=0$

    Two critical values ...

    $x=\dfrac{4 \pm 4\sqrt{2}}{3}$

    At $x=\dfrac{4+4\sqrt{2}}{3} \approx 3.21895$, $y \approx 11.8123$

    tangent line at that point is $y=11.8123$, as shown in the attached graph.

    How did you arrive at your tangent line equation?
    Attached Thumbnails Attached Thumbnails critical points-image.png  
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  5. #5
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    Re: critical points

    Thanks skeeter but that equation x^3 - 4x^2 - 5x + 36 should be x^3 - 4x^2 - 9x + 36.

    So derivative 3x^2 - 8x - 9.
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  6. #6
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    Re: critical points

    Ok ... so the criticsl values are $x=\dfrac{4 \pm \sqrt{43}}{3}$

    at $x=\dfrac{4+\sqrt{43}}{3} \approx 3.51915$, $y \approx -1.6274$

    Same drill ... tangent line is $y=-1.6274$. new graph attached

    Next time post the function and what you were supposed to find (i.e., directions) ... the thumbnail you posted isn't that large even after clicking on it.
    Attached Thumbnails Attached Thumbnails critical points-image.png  
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    Re: critical points

    setting the derivative = 0 I get x = 3.49?
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    Re: critical points

    Quote Originally Posted by anthonye View Post
    setting the derivative = 0 I get x = 3.49?
    How did you arrive at that value?
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  9. #9
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    Re: critical points

    Ok

    3x^2 - 8x - 9 = 0

    3x^2 - 8x = 9

    x^2 - 8/3x = 3

    x^2 - 8/3x + 1.8 = 4.8

    (x - 1.3)(x - 1.3) = 4.8

    (x - 1.3)^2 = 4.8

    (x - 1.3) = + or - 2.19

    x = 2.19 + 1.3 = 3.49
    x = 2.19 - 1.3 = -0.89.
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  10. #10
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    Re: critical points

    Note that $\left(\dfrac{4}{3}\right)^2 = \dfrac{16}{9} \ne 1.8$. ... your calculation is off because you rounded too soon. Put the calculator away until the very end ...


    $x^2 - \dfrac{8}{3}x = 3$

    $x^2 - \dfrac{8}{3}x + \left(\dfrac{4}{3}\right)^2 = 3+\left(\dfrac{4}{3}\right)^2$

    $\left(x-\dfrac{4}{3}\right)^2 = \dfrac{43}{9}$

    $x-\dfrac{4}{3} = \pm \dfrac{\sqrt{43}}{3}$

    $x = \dfrac{4 \pm \sqrt{43}}{3}$

    NOW use the calculator ...
    Last edited by skeeter; Sep 10th 2016 at 06:45 AM.
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  11. #11
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    Re: critical points

    Ok thanks skeeter its correct now, thanks for all your help.
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