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Math Help - Instantaneous rate of change?

  1. #1
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    Instantaneous rate of change?

    I don't know how to do this.. But f(x) = x + lnx..

    WHat's the value of c for which the instantaneous rate of change of f at x=c is the same as the average rate of change of f over [1, 4]?

    What do I do? I did the mean value theorem, but I didn't get the correct answer.. Do I take the derivative and then plug in 1 or 4 or what??
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by elpermic View Post
    I don't know how to do this.. But f(x) = x + lnx..

    WHat's the value of c for which the instantaneous rate of change of f at x=c is the same as the average rate of change of f over [1, 4]?

    What do I do? I did the mean value theorem, but I didn't get the correct answer.. Do I take the derivative and then plug in 1 or 4 or what??
    Applying the mean value theorem, you need to find a c that satisfies the equation 1+\frac1c=\frac{3+\ln 4}{3}. Obviously, c\neq0.

    Can you take it from here?
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  3. #3
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    Quote Originally Posted by Chris L T521 View Post
    Applying the mean value theorem, you need to find a c that satisfies the equation 1+\frac1c=\frac{3+\ln 4}{3}. Obviously, c\neq0.

    Can you take it from here?
    sorry for posting in between someone else's query. but can u please elaborate on
    <br />
\frac{3+\ln 4}{3}<br />
I didnt get how u got this.

    thanks
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by champrock View Post
    sorry for posting in between someone else's query. but can u please elaborate on
    <br />
\frac{3+\ln 4}{3}<br />
I didnt get how u got this.

    thanks
    By the intermediate value theorem, f^{\prime}\!\left(c\right)=\frac{f\!\left(b\right)-f\!\left(a\right)}{b-a}.

    In this case, f^{\prime}\!\left(x\right)=1+\frac{1}{x}\implies f^{\prime}\!\left(c\right)=1+\frac{1}{c}

    Now, \frac{f\!\left(b\right)-f\!\left(a\right)}{b-a}\implies\frac{f\!\left(4\right)-f\!\left(1\right)}{4-1}=\frac{\left(4+\ln 4\right)-\left(1+\ln1\right)}{3}=\frac{3+\ln 4}{3}, since \ln 1=0.
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