What is the absolute extrema of f(x)=sin^2(2x). Intervals [0,pi]
clearly, since 0 <= |sin x| <= 1, the extrema are where sin^2(2x) = 1 or 0
That is, sin(2x) = ±1 or sin(2x) = 0
In the indicated domain, that leaves
2x = π/2 or 3π/2 or 0 or π or 2π
x = π/4 or 3π/4 or 0 or π/2 or π
See the graph at
http://www.wolframalpha.com/input/?i=sin^2%282x%29
To find the absolute extrema of the function f(x) = sin^2(2x) on the interval [0, pi], we need to analyze the critical points and the endpoints of the interval.
1. Critical Points: These occur where the derivative of the function is either zero or undefined. First, let's find the derivative of f(x):
f'(x) = d/dx(sin^2(2x))
= 2sin(2x) * (cos(2x)) * 2
= 4sin(2x)cos(2x)
= 2sin(4x)
To find the critical points, we need to solve the equation f'(x) = 0. So, we set 2sin(4x) = 0 and solve for x:
sin(4x) = 0
4x = 0, pi, 2pi, 3pi, ...
This gives us x = 0, pi/4, pi/2, 3pi/4, ...
However, we need to check if these critical points are within the interval [0, pi]. From this list, only x = 0 and x = pi are within the given interval.
2. Endpoints: We also need to evaluate the function at the endpoints of the interval.
f(0) = sin^2(2(0)) = sin^2(0) = 0^2 = 0
f(pi) = sin^2(2(pi)) = sin^2(2pi) = sin^2(0) = 0^2 = 0
Now we have the function values at the critical points and the endpoints.
In summary:
- f(0) = 0
- f(pi) = 0
Since f(0) and f(pi) are both equal to 0, the absolute extrema of f(x) = sin^2(2x) on the interval [0, pi] are both 0.