Find any critical numbers of the function
h(x)=sin^2(x)+cos(x)
0 < x < 2π
h' = 2sin cos - sin
= sin(2cos - 1)
so, places where sinx = 0 or cosx = 1/2 are critical numbers.
piece o' cake, right?
To find the critical numbers of a function, we need to find the values of x where either the derivative is equal to zero or the derivative is undefined.
First, let's find the derivative of the function h(x) = sin^2(x) + cos(x). The derivative of sin^2(x) is 2sin(x)cos(x) using the chain rule, and the derivative of cos(x) is -sin(x). Therefore, the derivative of h(x) is:
h'(x) = 2sin(x)cos(x) - sin(x)
Now, we need to set h'(x) equal to zero and solve for x:
2sin(x)cos(x) - sin(x) = 0
Factoring out sin(x), we get:
sin(x)(2cos(x) - 1) = 0
This equation is satisfied when sin(x) = 0 or 2cos(x) - 1 = 0.
For sin(x) = 0, the critical numbers occur at x = 0, π, and 2π since sin(x) = 0 at these values within the given domain (0 < x < 2π).
For 2cos(x) - 1 = 0, we solve for x:
2cos(x) - 1 = 0
2cos(x) = 1
cos(x) = 1/2
The solutions for cos(x) = 1/2 within the given domain (0 < x < 2π) are x = π/3 and x = 5π/3.
Therefore, the critical numbers of the function h(x) = sin^2(x) + cos(x) within the domain 0 < x < 2π are: x = 0, π, 2π, π/3, and 5π/3.