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Finding the value guaranteed by the Mean Value Theorem

Consider the function f(x)=1xf(x) = \dfrac{1}{x} on the closed interval [1,3][1,3].

The function f(x)=1xf(x) = \dfrac{1}{x} is continuous on [1,3][1,3] and differentiable on (1,3)(1,3), so the Mean Value Theorem guarantees the existence of at least one number cc in (1,3)(1,3) such that

f′(c)=f(3)−f(1)3−1.f'(c) = \frac{f(3)-f(1)}{3-1}.

Find the value of cc that satisfies this equation.

Hint

First compute the average rate of change f(3)−f(1)3−1\frac{f(3)-f(1)}{3-1}. Then find f′(x)f'(x), set f′(c)f'(c) equal to that average rate of change, and solve for cc, keeping only the solution that lies in (1,3)(1,3).

Step-by-step solution

Compute the average rate of change: f(3)=13f(3) = \frac13, f(1)=1f(1) = 1, so f(3)−f(1)3−1=13−12=−232=−13\dfrac{f(3)-f(1)}{3-1} = \dfrac{\frac13 - 1}{2} = \dfrac{-\frac23}{2} = -\frac13.

Since f(x)=x−1f(x) = x^{-1}, we have f′(x)=−x−2=−1x2f'(x) = -x^{-2} = -\dfrac{1}{x^2}. Setting f′(c)=−13f'(c) = -\dfrac13 gives −1c2=−13-\dfrac{1}{c^2} = -\dfrac13, so c2=3c^2 = 3, meaning c=±3c = \pm\sqrt{3}. Since cc must lie in (1,3)(1,3), the valid solution is c=3c = \sqrt{3}.

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