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Intervals of increase/decrease and classification of local extrema

Consider the function r(x)=2sin⁡(x)+sin⁡(2x)r(x) = 2\sin(x) + \sin(2x) on the interval [0,2π][0, 2\pi].

Find the critical numbers of r(x)=2sin⁡(x)+sin⁡(2x)r(x) = 2\sin(x) + \sin(2x) in [0,2π][0, 2\pi], determine the intervals on which rr is increasing and decreasing, and use the First Derivative Test to classify each critical point as a local maximum, local minimum, or neither.

Hint

Compute r′(x)=2cos⁡(x)+2cos⁡(2x)r'(x) = 2\cos(x) + 2\cos(2x), use the double angle identity cos⁡(2x)=2cos⁡2(x)−1\cos(2x) = 2\cos^2(x) - 1 to rewrite r′(x)r'(x) as a quadratic in cos⁡(x)\cos(x), factor to find where r′(x)=0r'(x) = 0 on [0,2π][0, 2\pi], then check the sign of r′r' on each subinterval.

Step-by-step solution

First compute r′(x)=2cos⁡(x)+2cos⁡(2x)r'(x) = 2\cos(x) + 2\cos(2x). Using cos⁡(2x)=2cos⁡2(x)−1\cos(2x) = 2\cos^2(x) - 1, rewrite this as r′(x)=4cos⁡2(x)+2cos⁡(x)−2=2(2cos⁡(x)−1)(cos⁡(x)+1)r'(x) = 4\cos^2(x) + 2\cos(x) - 2 = 2(2\cos(x) - 1)(\cos(x)+1). Setting r′(x)=0r'(x) = 0 gives cos⁡(x)=1/2\cos(x) = 1/2 or cos⁡(x)=−1\cos(x) = -1. On [0,2π][0, 2\pi] this yields x=π/3,π,5π/3x = \pi/3, \pi, 5\pi/3 as critical points.

Testing sample points in each subinterval: on (0,π/3)(0, \pi/3) pick x=π/6x=\pi/6: r′(π/6)=2(2cos⁡(π/6)−1)(cos⁡(π/6)+1)=2(3−1)(32+1)>0r'(\pi/6) = 2(2\cos(\pi/6) - 1)(\cos(\pi/6)+1) = 2(\sqrt{3} - 1)(\frac{\sqrt{3}}{2}+1) > 0 (increasing). On (π/3,π)(\pi/3, \pi) pick x=π/2x=\pi/2: r′(π/2)=2(2⋅0−1)(0+1)=−2<0r'(\pi/2) = 2(2\cdot 0 - 1)(0 + 1) = -2 < 0 (decreasing). On (π,5π/3)(\pi, 5\pi/3) pick x=3π/2x=3\pi/2: r′(3π/2)=2(2⋅0−1)(0+1)=−2<0r'(3\pi/2) = 2(2\cdot 0 - 1)(0 + 1) = -2 < 0 (decreasing). On (5π/3,2π)(5\pi/3, 2\pi) pick x=11π/6x=11\pi/6: r′(11π/6)=2(2cos⁡(11π/6)−1)(cos⁡(11π/6)+1)=2(3−1)(32+1)>0r'(11\pi/6) = 2(2\cos(11\pi/6) - 1)(\cos(11\pi/6)+1) = 2(\sqrt{3} - 1)(\frac{\sqrt{3}}{2}+1) > 0 (increasing).

Intervals of increase/decrease and classification of local extrema — solution graph

To find the local extrema values: at x=π/3x = \pi/3, r(π/3)=2sin⁡(π/3)+sin⁡(2π/3)=2⋅32+32=332r(\pi/3) = 2\sin(\pi/3) + \sin(2\pi/3) = 2 \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{2} = \frac{3\sqrt{3}}{2}. At x=5π/3x = 5\pi/3, r(5π/3)=2sin⁡(5π/3)+sin⁡(10π/3)=2⋅(−32)+(−32)=−332r(5\pi/3) = 2\sin(5\pi/3) + \sin(10\pi/3) = 2 \cdot (-\frac{\sqrt{3}}{2}) + (-\frac{\sqrt{3}}{2}) = -\frac{3\sqrt{3}}{2}. Since r′r' changes from positive to negative at x=π/3x = \pi/3, this is a local maximum. Since r′r' changes from negative to positive at x=5π/3x = 5\pi/3, this is a local minimum. At x=πx = \pi, r′(π)=0r'(\pi) = 0 but r′r' remains negative on both (π/3,π)(\pi/3, \pi) and (π,5π/3)(\pi, 5\pi/3), so r′r' does not change sign at this critical point, making it neither a local maximum nor minimum.

Behavior at the critical point where the derivative fails to change sign

Hint
Step-by-step solution

At x = π, the derivative equals zero since cos(π) = -1 makes the factor (cos(x) + 1) equal zero. However, the derivative does not change sign at this point. Since r'(x) = 2(2cos(x) - 1)(cos(x) + 1) is negative immediately before π (where cos(x) > -1) and remains negative immediately after π (where cos(x) > -1), the First Derivative Test cannot classify this as a local extremum. The function is decreasing through x = π with no turning behavior. Therefore this critical point is neither a local maximum nor a local minimum.

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