Answer:
A) Increasing from 0 to π/4
Decreasing from π/4 to π/2
B) f(x) is concave upwards on the interval, π/4 < x < (3/4)·π
f(x) is concave downwards on the interval, 0 < x < π/2
C) Where n = 0, 1, 2, ..., the inflection points are x = π/8 + n·π/2
Explanation:
A) The given function is, f(x) = sin²2·x, (0, π)
We note that at x = 0, f(0) = sin²(2×0) = 0
The maximum value of the sine function is sin(π/2), therefore, at the maximum value, we have;
2·x = π/2
x = π/4
At x = π/4, we have;
The maximum value, f(π/4) = sin²(2×π/4) = 1
At x = π/8, we have; f(π/8) = sin²(2×π/8) = 0.5
Using sign analysis, we have;
(f(π/4) - f(0))/(1 - 0) = (1 - 0)/(π/4 - 0) ≈ 1.27 (The slope is positive between 0 and π/4) and therefore, the function is increasing from 0 to π/4
Between π/4 and π/2, we have;
f(π/2) = sin²(2×π/2) = 0
Therefore, the slope, m = (0 - 1)/(π/2- π/4) ≈ -1.27 (The slope is negative between π/4 and π/2) and therefore, the function is decreasing from π/4 to π/2
B) The function is concave up when the slope is negative to the left of a point and positive to the right of the same point
Taking the point π/2, we have;
To the left, x = π/4, and the slope is (0 - 1)/(π/2- π/4) ≈ -1.27 is negative
To the right, x = (3/4)·π, f((3/4)·π) = sin²((3/4)·π) = 1, therefore;
The slope = (1 - 0)/((3/4)·π - π/2) ≈ 1.27, the slope is positive
Therefore, in the interval π/4 < x < (3/4)·π, the curve is concave upwards
The concave down where the slope to the left of a point is positive (the function is increasing) and the slope to the right is negative (the function is decreasing)
Therefore, the function is concave down in the interval, 0 < x < π/2
C) The inflection points are the turning points which are the points across which the slope changes sign
At the inflection point, we have;
f''(x) = d²(sin²2·x)/dx = 8·cos(4·x) = 0
∴ cos(4·x) = 0
4·x = arccos(0) = π/2
∴ The inflection points x = π/8 + n·π/2