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Thread: Continuity Functions

  1. #1
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    Continuity Functions

    I'm at a loss on where to start with either of these problems. It seems you will use the same basic methods of solving each so I will list each of them.

    1.) For what value of the constant is the function continuous on where


    2.) For what value of is the function defined below continuous on ?


    Thanks for help in advance.
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  2. #2
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    Hello, jwebb19!

    1.) For what value of the constant $\displaystyle c$ is the function $\displaystyle f$ continuous?

    . . $\displaystyle f(t) \;=\;\begin{Bmatrix}t^2-c,& t < 6 \\ ct + 3, & t \geq 6\end{Bmatrix} $

    To the left of $\displaystyle t = 6$, the graph is a parabola -- continuous.
    To the right of $\displaystyle t = 6$, the graph is a straight line -- continuous.

    The only discontinuity would be at $\displaystyle t = 6$.
    To be a continuous function, the two curves must "meet" at $\displaystyle t = 6.$

    So we have: .$\displaystyle f(4) \;=\;36-c \:=\:6c + 3\quad\Rightarrow\quad 7c \:=\:33$

    Therefore: .$\displaystyle c\,=\,\frac{33}{7}$



    2.) For what value of $\displaystyle c$ is the function $\displaystyle f$ continuous?

    . . $\displaystyle f(x) \;=\;\begin{Bmatrix}cx + 3, & x < 4 \\ cx^2-3, & x \geq 4\end{Bmatrix}$

    The two branches of the function must be equal at $\displaystyle x = 4.$

    Hence: .$\displaystyle 4c + 3 \:=\:16c - 3\quad\Rightarrow\quad12c \:=\:6$

    Therefore: .$\displaystyle c\,=\,\frac{1}{2}$

    Last edited by Soroban; Sep 11th 2007 at 10:49 PM.
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  3. #3
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    Thanks for the great explanation. Clarified everything. Thanks again.
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