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Finding the MVT guaranteed point for arctan(x)

Consider the function f(x)=arctan⁡(x)f(x) = \arctan(x) on the closed interval [0,3][0, \sqrt{3}].

The function f(x)=arctan⁡(x)f(x) = \arctan(x) is continuous on [0,3][0, \sqrt{3}] and differentiable on (0,3)(0, \sqrt{3}), so the Mean Value Theorem guarantees at least one number cc in (0,3)(0, \sqrt{3}) such that f′(c)=f(3)−f(0)3−0.f'(c) = \frac{f(\sqrt{3}) - f(0)}{\sqrt{3} - 0}. Find the exact value of cc that satisfies this equation.

Hint

First compute f(3)f(\sqrt{3}) and f(0)f(0) to get the slope of the secant line. Then set f′(c)=11+c2f'(c) = \frac{1}{1+c^2} equal to that slope and solve for cc, keeping only the solution in (0,3)(0, \sqrt{3}).

Step-by-step solution

First find the average rate of change: f(3)=arctan⁡(3)=π/3f(\sqrt{3}) = \arctan(\sqrt{3}) = \pi/3 and f(0)=arctan⁡(0)=0f(0) = \arctan(0) = 0, so the secant slope is π/3−03−0=π33.\frac{\pi/3 - 0}{\sqrt{3} - 0} = \frac{\pi}{3\sqrt{3}}. This is the value that f′(c)f'(c) must equal.

Since f′(x)=11+x2f'(x) = \frac{1}{1+x^2}, we set 11+c2=π33\frac{1}{1+c^2} = \frac{\pi}{3\sqrt{3}}. Solving gives 1+c2=33π1 + c^2 = \frac{3\sqrt{3}}{\pi}, so c2=33π−1c^2 = \frac{3\sqrt{3}}{\pi} - 1, and taking the positive root (since cc must lie in (0,3)(0,\sqrt{3})) gives c=33π−1c = \sqrt{\frac{3\sqrt{3}}{\pi} - 1}.

Geometric meaning of the guaranteed point

Hint
Step-by-step solution

They are parallel. The Mean Value Theorem guarantees a point where the tangent line's slope equals the secant line's slope connecting the endpoints, so the two lines never intersect and run in the same direction.

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