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Thread: Mean Value theorem question

  1. #1
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    Mean Value theorem question

    Question: Verify that the function satisfies the hypotheses of The Mean Value Theorem on the given interval and find all c that satisfy the conclusion of the MTV.

    $\displaystyle f(x)= \frac{x}{x+2}$, [1,4]
    ---

    -continuous and differentiable
    f(1)=1/3
    f(4)=2/3

    $\displaystyle \frac{2/3-1/3}{4-1}$

    $\displaystyle \frac{1/3}{3}$
    =1/9

    $\displaystyle
    f'=\frac{(x+2) -x}{(x+2)^2}
    $

    $\displaystyle
    f'=\frac{(2)}{(x+2)^2}
    $=1/9

    =$\displaystyle (x+2)^2=18$
    =$\displaystyle (x^2+4x-14)$

    Did I make a mistake or am I suppose to use the quadratic formula to get c?
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  2. #2
    Junior Member utopiaNow's Avatar
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    Quote Originally Posted by hazecraze View Post
    Question: Verify that the function satisfies the hypotheses of The Mean Value Theorem on the given interval and find all c that satisfy the conclusion of the MTV.

    $\displaystyle f(x)= \frac{x}{x+2}$, [1,4]
    ---

    -continuous and differentiable
    f(1)=1/3
    f(4)=2/3

    $\displaystyle \frac{2/3-1/3}{4-1}$

    $\displaystyle \frac{1/3}{3}$
    =1/9

    $\displaystyle
    f'=\frac{(x+2) -x}{(x+2)^2}
    $

    $\displaystyle
    f'=\frac{(2)}{(x+2)^2}
    $=1/9

    =$\displaystyle (x+2)^2=18$
    =$\displaystyle (x^2+4x-14)$

    Did I make a mistake or am I suppose to use the quadratic formula to get c?
    Looks fine, yes use the quadratic formula to find the $\displaystyle c$ in the interval $\displaystyle a < c < b$ ie. accept the $\displaystyle c$ in this interval and reject the value for $\displaystyle c $ which is not in this interval.
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  3. #3
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    Quote Originally Posted by hazecraze View Post
    Question: Verify that the function satisfies the hypotheses of The Mean Value Theorem on the given interval and find all c that satisfy the conclusion of the MTV.

    $\displaystyle f(x)= \frac{x}{x+2}$, [1,4]
    ---

    -continuous and differentiable
    f(1)=1/3
    f(4)=2/3

    $\displaystyle \frac{2/3-1/3}{4-1}$

    $\displaystyle \frac{1/3}{3}$
    =1/9

    $\displaystyle
    f'=\frac{(x+2) -x}{(x+2)^2}
    $

    $\displaystyle
    f'=\frac{(2)}{(x+2)^2}
    $=1/9

    =$\displaystyle (x+2)^2=18$
    =$\displaystyle (x^2+4x-14)$

    Did I make a mistake or am I suppose to use the quadratic formula to get c?

    So far so good, now continue: $\displaystyle (x+2)^2=18 \Longrightarrow x+2=\pm \sqrt{18}=\pm 3\sqrt{2} \Longrightarrow$$\displaystyle x=\left\{\begin{array}{cc}3\sqrt{2}-2\\-3\sqrt{2}-2\end{array}\right.$

    Of course, you need (and can) take only one of the above values.

    Tonio
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