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Classifying a critical point using the Second Derivative Test

Consider the function f(x)=cos⁡(2x)−2sin⁡(x)f(x) = \cos(2x) - 2\sin(x) on the interval [0,2π][0, 2\pi].

The critical numbers of f(x)=cos⁡(2x)−2sin⁡(x)f(x) = \cos(2x) - 2\sin(x) on [0,2π][0, 2\pi] occur where f′(x)=−2sin⁡(2x)−2cos⁡(x)=0f'(x) = -2\sin(2x) - 2\cos(x) = 0. One of these critical numbers is x=π2x = \dfrac{\pi}{2}. Use the Second Derivative Test to classify ff at x=π2x = \dfrac{\pi}{2} as a local maximum, a local minimum, or neither. State your classification and justify it using the value of f′′(π2)f''\left(\frac{\pi}{2}\right).

Hint

First find f′′(x)f''(x) by differentiating f′(x)=−2sin⁡(2x)−2cos⁡(x)f'(x) = -2\sin(2x) - 2\cos(x). Then plug in x=π/2x = \pi/2 and check the sign of the result: positive means concave up (local min), negative means concave down (local max).

Step-by-step solution

Differentiate f′(x)=−2sin⁡(2x)−2cos⁡(x)f'(x) = -2\sin(2x) - 2\cos(x) to get f′′(x)=−4cos⁡(2x)+2sin⁡(x)f''(x) = -4\cos(2x) + 2\sin(x). This uses the chain rule on sin⁡(2x)\sin(2x) and the derivative of cos⁡(x)\cos(x).

Evaluate at x=π/2x = \pi/2: f′′(π/2)=−4cos⁡(π)+2sin⁡(π/2)=−4(−1)+2(1)=4+2=6f''(\pi/2) = -4\cos(\pi) + 2\sin(\pi/2) = -4(-1) + 2(1) = 4 + 2 = 6. Since f′′(π/2)=6>0f''(\pi/2) = 6 > 0, the function is concave up at this point, so by the Second Derivative Test ff has a local minimum at x=π/2x = \pi/2.

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