Find the absolute maximum and minimum values of f on the given closed interval, and state where those values occur.

f(x)=5x-10sinx ; [-pi/4,pi/2]

To find the absolute maximum and minimum values of f(x) on the closed interval [-π/4, π/2], we can follow these steps:

1. Find the critical points of f(x) within the interval [-π/4, π/2]. These points occur where the derivative of f(x) is equal to zero or does not exist.

2. Evaluate the function f(x) at the critical points and endpoints of the interval.

3. Compare the values obtained in step 2 to identify the absolute maximum and minimum.

Now let's proceed with these steps:

Step 1: Find the critical points
To find the critical points, we need to find where the derivative of f(x) is equal to zero or undefined. Let's calculate the derivative of f(x):

f'(x) = 5 - 10cos(x)

Now, set f'(x) = 0 and solve for x:

5 - 10cos(x) = 0
10cos(x) = 5
cos(x) = 1/2
x = π/3, 5π/3

Step 2: Evaluate f(x) at critical points and endpoints
Now, we need to evaluate f(x) at x = -π/4, x = π/2, x = π/3, and x = 5π/3:

f(-π/4) = 5(-π/4) - 10sin(-π/4) = -5π/4 + 10/√2 ≈ -3.511
f(π/2) = 5(π/2) - 10sin(π/2) = 5π/2 - 10 ≈ 9.425
f(π/3) = 5(π/3) - 10sin(π/3) = 5π/3 - 10√3/2 ≈ -1.108
f(5π/3) = 5(5π/3) - 10sin(5π/3) = -5π/3 - 10√3/2 ≈ -3.108

Step 3: Determine absolute maximum and minimum
Comparing the values obtained in step 2, we find:
Absolute maximum value: f(π/2) ≈ 9.425 at x = π/2
Absolute minimum value: f(5π/3) ≈ -3.108 at x = 5π/3

Therefore, the absolute maximum value of f(x) on the interval [-π/4, π/2] is approximately 9.425, which occurs at x = π/2. The absolute minimum value is approximately -3.108, which occurs at x = 5π/3.