Find monotonic rising interval of an exponential function

Question:

If function y = 2^{sin2x}, find its monotonic rising interval.

Solution:

Because the function y = 2^{x} is a monotonic rising function. So, the monotonic rising interval of y = 2^{sin2x} is the monotonic rising interval of y = sin2x.

Now we are going to find the monotonic rising interval of the function y = sin2x. Look the graph of y = sin x, its monotonic rising interval is:

2pi k - pi/2 <= x <= pi/2 + 2pi k

So, the monotonic rising interval of the function y = sin2x is:

2pi k - pi/2 <= 2x <= pi/2 + 2pi k

pi k - pi/4 <= x < pi/4 + pi k

when k = 0, its monotonic rising interval is: -pi/4 <= x <= pi/4

when k = 1, its monotonic rising interval is: pi - pi/4 <= x <= pi/4 + pi, which is, 3pi/4 <= x <= 5pi/4

The graph above is y = 2^{sin2x}. Its first monotonic rising interval is from x = -pi/4 to x = pi/4. Itâ€™s the second rising interval is from x = 3pi/4 to 5pi/4.

when x = -pi/4, y = 2^{sin2(-pi/4)} = 2^{sin(-pi/2)} = 2^{-sin(pi/2)} = 2^{-1} = 1/2 = 0.5

when x = pi/4, y = 2^{sin2(pi/4)} = 2^{sin(pi/2)} = 2^{1} = 2

when x = 3pi/4, y = 2^{sin2(3pi/4)} = 2^{sin(3pi/2)} = 2^{-1} = 1/2 = 0.5

when x = 5pi/4, y = 2^{sin2(5pi/4)} = 2^{sin(5pi/2)} = 2^{sin(2pi + pi/2)} = 2^{sin(pi/2)} = 2^{1} = 2.