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Math Help - Piecewise function

  1. #1
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    Piecewise function

    Piecewise function-piecewise.jpg
    For the function f(x) = { -3x + 2 if x < 1
    { x^2 if x > 1

    a) sketch the piecewise function

    b) determine the x-values, if any, at which the function is discontinuous. Find appropriate limits to support your conclusion.




    Attached is my sketch, I don't know if it's correct?
    I don't know how to do b.

    Thank you in advanced.
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  2. #2
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    Quote Originally Posted by advancedfunctions2010 View Post
    Click image for larger version. 

Name:	piecewise.jpg 
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    For the function f(x) = { -3x + 2 if x < 1
    { x^2 if x > 1

    a) sketch the piecewise function

    b) determine the x-values, if any, at which the function is discontinuous. Find appropriate limits to support your conclusion.

    Attached is my sketch, I don't know if it's correct?
    I don't know how to do b.

    Thank you in advanced.
    The given function is: f(x)=\left\{\begin{array}{rcl}-3x+2&if&x<1 \\ x^2&if&x\geq 1\end{array}\right.

    1. The first term belongs to a linear function which is continuous. The 2nd term belongs to a quadratic function which is continuous. The only possible value at which discontinuity could occur is x = 1.

    2. Determine

    \lim_{x \to 1 \wedge x<1}(-3x+2)

    and

    \lim_{x \to 1 \wedge x\geq1}(x^2)
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  3. #3
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    Hello, advancedfunctions2010!

    Your sketch is wrong . . .


    For the function: . f(x) \:=\:\begin{Bmatrix}-3x + 2 && x < 1 \\ x^2 && x \ge 1 \end{array}

    a) Sketch the piecewise function

    b) Determine the x-values, if any, at which the function is discontinuous.
    Find appropriate limits to support your conclusion.
    Code:
            *     |          *
             *    |
              *   |         *
               *  |        *
                * |      *
                 *|   *
                  *
                  |*  1
          - - - - + * + - - - -
                  |  *
                  |   o
                  |

    Obviously there is a discontinuity at x = 1.

    We verify this by taking the limit of f(x) "from both sides".


    Take limit of f(x) as \,x approaches 1 "from below".

    . . \displaystyle \lim_{x\to1^-}f(x) \:=\:\lim_{x\to1^-}(-3x+2) \;=\;-1


    Take the limit of f(x) as \,x approaches 1 "from above".

    . . \displaystyle \lim_{x\to1^+} f(x) \;=\;\lim_{x\to1^+}x^2 \;=\;1


    Since the two limits are not equal, f(x) is discontinuous at x = 1.


    Edit: Too slow again . . . redundant explanation.
    .
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