Inactive Tutor answered • 06/10/23
We can start by finding all critical points of h(x) on this interval. Note that h(x) is differentiable on the whole interval and h'(x) = 1 + 2 sin(x).
To find critical points, set h'(x) = 1 + 2 sin(x) = 0, so sin(x) = -1/2. On the interval [0, 4π], the solutions are x = 7π/6, 11π/6, 19π/6, and 23π/6.
Next, we examine the second derivative h''(x) = 2 cos(x) at each of these points.
h''(7π/6) = h''(19π/6) = - √3 < 0.
h''(11π/6) = h''(23π/6) = √3 > 0.
By the second derivative test, we know that if h'(c) = 0 and h''(c) < 0, then h has a local maximum at x = c. In our example, the local minimums are at x = 7π/6 and x = 19π/6. If we plug these x-values into h(x), we find h(7π/6) = 7π/6 + √3 and h(19π/6) = 19π/6 + √3.
Putting it all together, the local maxima of h(x) on the interval [0, 4π] are (7π/6, 7π/6 + √3) and (19π/6, 19π/6 + √3).
Notice that we know our other two critical points are not maxima because h''(x) evaluated at these points is positive. These two points are actually local minima.
