find absolute max and min of f(x)=cosx+sin^2(x) on interval [0,2pi]. Give exact answers, not decimal approx. (Use closed interval method)
f(0) = 1+0 = 1
f(2pi) = 1+0 = 1
f'(x) = -sinx + 2sinx cosx
= sinx(2cosx-1)
So, f'=0 at x=0,pi,2pi and pi/3,5pi/3
f(pi/3) = 1/2 + 3/4 = 5/4
f(pi) = -1+0 = -1
f(5pi/3) = 1/2 + 3/4 = 5/4
Looks like the min is -1 and the max is 5/4
See
http://www.wolframalpha.com/input/?i=cosx%2Bsin^2%28x%29
To find the absolute maximum and minimum of the function f(x) = cos(x) + sin²(x) on the interval [0, 2π] using the closed interval method, follow these steps:
1. Determine the critical points of the function within the interval [0, 2π]. Critical points are the values of x where the derivative of f(x) is either zero or undefined.
To find the derivative of f(x), apply the chain rule:
f'(x) = -sin(x) + 2sin(x)cos(x)
Setting f'(x) = 0 to find the critical points:
-sin(x) + 2sin(x)cos(x) = 0
sin(x)(2cos(x) - 1) = 0
This equation implies that either sin(x) = 0 or 2cos(x) - 1 = 0.
For sin(x) = 0, the solutions are x = 0 and x = π.
For 2cos(x) - 1 = 0, the solution is x = π/3.
2. Evaluate the function f(x) at the critical points and at the endpoints of the interval [0, 2π].
f(0) = cos(0) + sin²(0) = 1 + 0 = 1
f(π/3) = cos(π/3) + sin²(π/3) = 1/2 + (sqrt(3)/2)² = 1/2 + 3/4 = 5/4
f(π) = cos(π) + sin²(π) = -1 + 0 = -1
f(2π) = cos(2π) + sin²(2π) = 1 + 0 = 1
3. Compare the function values obtained at the critical points and endpoints to find the absolute maximum and minimum.
From the evaluations in step 2, we have:
f(0) = 1
f(π/3) = 5/4
f(π) = -1
f(2π) = 1
The absolute maximum is 1, which occurs at x = 0 and x = 2π.
The absolute minimum is -1, which occurs at x = π.
Therefore, the absolute maximum of f(x) on the interval [0, 2π] is 1, and the absolute minimum is -1.