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Math Help - Finding crritical points of a piecewise function

  1. #1
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    Finding crritical points of a piecewise function

    Given:

    f(x)=\left\{\begin{array}{cc}-x^3+1 ,&\mbox{ if }<br />
x\leq 0\\-x^2 +2x, & \mbox{ if } x>0\end{array}\right.

    Find the critical numbers and open intervals on which the function is increasing or decreasing.

    Would I simply compute the derivative of -x^3+1 and -x^2+2x seperately and solve each down?

    Can anyone offer some instruction for piecewise functions ?
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  2. #2
    No one in Particular VonNemo19's Avatar
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    Quote Originally Posted by jimmyp View Post
    Given:

    f(x)=\left\{\begin{array}{cc}-x^3+1 ,&\mbox{ if }<br />
x\leq 0\\-x^2 +2x, & \mbox{ if } x>0\end{array}\right.

    Find the critical numbers and open intervals on which the function is increasing or decreasing.

    Would I simply compute the derivative of -x^3+1 and -x^2+2x seperately and solve each down?

    Can anyone offer some instruction for piecewise functions ?
    yes

    all youre trying to do is find the same things you would in a regular function.

    sketch the graph and then estimate where the critical values are. see if this is a good approximation of doing it analytically
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  3. #3
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    Thus:

    y' = -3x^2
    and
    -2x+2


    -3x^2=0
    x=0

    and:

    -2x+2=0
    x=1

    Critical point: (0,1)


    And the test intervals are: (- \infty, 0),(1, \infty)

    Thus:

    - \infty < x < 0
    test value: -2
    Sign of f'(x) on -2x+2 = -2
    Sign of f'(x) on -3x^2 = -12
    Conclusion: decreasing for x < 0 and decreasing for x > 0

    and:

    \infty < x < 1
    test value: 2
    Sign of f'(x) on -2x+2 = -2
    Sign of f'(x) on -3x^2 = -12
    Conclusion: decreasing for x < 0 and decreasing for x > 0


    is this correct or have i made a mistake ?
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  4. #4
    Senior Member pankaj's Avatar
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    The intervals of increasing monotonicity are (0,1) and the interval where f(x) decreases are (-\infty,0)\cup (1,\infty)

    You forgot to check for the sign of derivative on the interval (0,1)
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