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Applying the Mean Value Theorem to ln(x)

Consider the function f(x)=ln⁡(x)f(x) = \ln(x) on the closed interval [1,e][1, e].

The function f(x)=ln⁡(x)f(x) = \ln(x) is continuous on [1,e][1, e] and differentiable on (1,e)(1, e), so the Mean Value Theorem guarantees the existence of at least one number cc in (1,e)(1, e) such that

f′(c)=f(e)−f(1)e−1.f'(c) = \frac{f(e) - f(1)}{e - 1}.

Find the exact value of cc that satisfies this equation.

Hint

First compute f(e)f(e) and f(1)f(1) to find the average rate of change (the slope of the secant line). Then compute f′(x)f'(x), set f′(c)f'(c) equal to that slope, and solve for cc.

Step-by-step solution

Since f(x)=ln⁡(x)f(x) = \ln(x), we have f(e)=1f(e) = 1 and f(1)=0f(1) = 0. The average rate of change over [1,e][1,e] is f(e)−f(1)e−1=1−0e−1=1e−1\dfrac{f(e)-f(1)}{e-1} = \dfrac{1-0}{e-1} = \dfrac{1}{e-1}.

The derivative is f′(x)=1xf'(x) = \dfrac{1}{x}. Setting f′(c)=1e−1f'(c) = \dfrac{1}{e-1} gives 1c=1e−1\dfrac{1}{c} = \dfrac{1}{e-1}, so c=e−1c = e - 1. Since 1<e−1<e1 < e-1 < e, this value lies in the required open interval, confirming it satisfies the MVT.

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