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Finding the value guaranteed by the Mean Value Theorem

Consider the function f(x)=cos⁡(x)f(x) = \cos(x) on the closed interval [−π2,π2]\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right].

Since f(x)=cos⁡(x)f(x) = \cos(x) is continuous on [−π2,π2]\left[-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right] and differentiable on (−π2,π2)\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), the Mean Value Theorem guarantees at least one value cc in (−π2,π2)\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) such that f′(c)=f(π/2)−f(−π/2)π/2−(−π/2)f'(c) = \dfrac{f(\pi/2) - f(-\pi/2)}{\pi/2 - (-\pi/2)}. Find the exact value of cc.

Hint

First compute the slope of the secant line using the endpoint values of ff. Then set f′(c)f'(c) equal to that slope and solve for cc, keeping in mind that cc must lie strictly inside the given interval.

Step-by-step solution

First evaluate the endpoints: f(π/2)=cos⁡(π/2)=0f(\pi/2) = \cos(\pi/2) = 0 and f(−π/2)=cos⁡(−π/2)=0f(-\pi/2) = \cos(-\pi/2) = 0. So the average rate of change is 0−0π/2−(−π/2)=0π=0\dfrac{0 - 0}{\pi/2 - (-\pi/2)} = \dfrac{0}{\pi} = 0. This means the secant line connecting the endpoints is horizontal.

Since f′(x)=−sin⁡(x)f'(x) = -\sin(x), the MVT requires −sin⁡(c)=0-\sin(c) = 0, so sin⁡(c)=0\sin(c) = 0. The only solution to this in the open interval (−π2,π2)\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right) is c=0c = 0, which corresponds to the point where the tangent line is horizontal, matching the horizontal secant line.

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