Mathematics

Problem

Let

k\ne 0

and

f(x)=e^{kx}

[0,1]

. Which expression gives the value of

c

(0,1)

where the instantaneous rate of change of

f

equals its average rate of change over the interval? a)

\dfrac{1}{k}\,(e^{k}-1)

\ln\!\left(\dfrac{e^{k}-1}{k}\right)

\dfrac{1}{k}\,\ln\!\left(\dfrac{e^{k}-1}{k}\right)

\dfrac{1}{2}

Intuitive explanation

This is the Mean Value Theorem in action—finding where the instantaneous rate of change (the derivative) matches the average rate of change (the slope of the secant line). For an exponential function, you'll need to set $f'(c)$ equal to $\frac{f(1)-f(0)}{1-0}$ , then solve for $c$ . The key insight is that exponential functions have a special property: their derivative is proportional to themselves, which makes the algebra cleaner than you might expect.

Intuitive explanation

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