Mathematics

Mathematics

Problem

For f(x)=ln⁡xf(x)=\ln x on the interval [1,e2][1,e^{2}], the average rate of change is 2e2−1\dfrac{2}{e^{2}-1}. At which value of xx in (1,e2)(1,e^{2}) does the instantaneous rate of change equal this average? a) e22\dfrac{e^{2}}{2} b) 1+e22\dfrac{1+e^{2}}{2} c) e2−12\dfrac{e^{2}-1}{2} d) ee
Intuitive explanation

The Mean Value Theorem guarantees that somewhere inside an interval, the instantaneous rate of change (derivative) must equal the average rate of change over the whole interval. Here you'll find that special point by setting f′(x)f'(x) equal to the given average value and solving for xx. Since f(x)=ln⁡xf(x) = \ln x, you already know the derivative—just equate it to 2e2−1\frac{2}{e^2-1} and isolate xx.

Intuitive explanation

The Mean Value Theorem guarantees that somewhere inside an interval, the instantaneous rate of change (derivative) must equal the average rate of change over the whole interval. Here you'll find that special point by setting f′(x)f'(x) equal to the given average value and solving for xx. Since f(x)=ln⁡xf(x) = \ln x, you already know the derivative—just equate it to 2e2−1\frac{2}{e^2-1} and isolate xx.

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