Arijit G. answered • 06/10/23
UC Berkeley Student Passionate About Math, Physics, and CS
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.
