f(x) = − cos(x^2) + 2sin(x)

How to find critical points, absolute max, and absolute min of f(x) on the interval [1,3.5]?

well, we already did the critical points (where f'(x) = 0)

so now it's just a matter of checking the sign of f"(x) at the critical points to see whether they are max or min.
f'(x) = 2x sin(x^2) + 2cosx
f"(x) = 4x^2 cos(x^2) + 2sin(x^2) - 2sinx

Come back with your work if you get stuck.

I think oobleck did this earlier today, but I can't find it, so ....

f(x) = − cos(x^2) + 2sin(x)
f '(x) = 2x sin(x^2) + 2cosx
No easy way to solve 2x sin(x^2) + 2cosx = 0 , which we have to do for your problem

Wolfram says: https://www.wolframalpha.com/input/?i=solve+2x+sin%28x%5E2%29+%2B+2cosx+%3D+0

I see 3 solutions that fit within your given domain

To find the critical points, absolute maximum, and absolute minimum of the function f(x) on the interval [1,3.5], first, we need to find the derivative of f(x) and set it equal to zero to find the critical points. Then, we will evaluate the function at the critical points and endpoints of the interval to determine the maximum and minimum values.

Step 1: Find the derivative of f(x)
The derivative of f(x) will help us locate the critical points. Take the derivative of each term separately using the chain rule and product rule when necessary.

f'(x) = (-1) * (-2x) * sin(x^2) + 2cos(x) = 2x * sin(x^2) + 2cos(x)

Step 2: Find the critical points
To find the critical points, set f'(x) equal to zero and solve for x.

2x * sin(x^2) + 2cos(x) = 0

There may not be a straightforward algebraic solution, so we can use numerical methods or calculators to approximate the solutions. One common method is Newton's method or using a graphing calculator that can find the zeros of a function.

Step 3: Evaluate the function at critical points and endpoints
Now, we need to evaluate the function at the critical points we found in Step 2, and at the endpoints of the interval [1,3.5].

Evaluate f(x) at the critical points: substitute the critical points (let's say x = a and x = b) back into the original function f(x).

f(a) = − cos(a^2) + 2sin(a)
f(b) = − cos(b^2) + 2sin(b)

Evaluate f(x) at the endpoints of the interval [1,3.5]: substitute the endpoints into the original function f(x).

f(1) = − cos(1^2) + 2sin(1)
f(3.5) = − cos(3.5^2) + 2sin(3.5)

Step 4: Determine the absolute maximum and minimum
By comparing the function values at the critical points and endpoints, we can find the absolute maximum and minimum.

- The highest value among f(a), f(b), f(1), and f(3.5) will be the absolute maximum.
- The lowest value among f(a), f(b), f(1), and f(3.5) will be the absolute minimum.

Note: If there are no critical points on the interval [1,3.5], we only need to compare the function values at the endpoints.

By following these steps, you will be able to find the critical points, absolute maximum, and absolute minimum of the function f(x) on the interval [1,3.5].