# Thread: Instantaneous rate of change?

1. ## 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??

2. Originally Posted by elpermic
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?

3. Originally Posted by Chris L T521
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
$
\frac{3+\ln 4}{3}
$
I didnt get how u got this.

thanks

4. Originally Posted by champrock
sorry for posting in between someone else's query. but can u please elaborate on
$
\frac{3+\ln 4}{3}
$
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 $\ln 1=0$.