find the value of the function at its absolute maximum and minimum values. let f be defined by the function…

Explanation:

Step1: Let (t = \cos x).

Since (0\leq x\leq\frac{3\pi}{2}), then (- 1\leq t\leq1). The function (y = f(x)=\cos^{2}x-\cos x) can be rewritten as (y=t^{2}-t).

Step2: Find the derivative of (y = t^{2}-t).

The derivative (y^\prime=2t - 1). Set (y^\prime = 0), then (2t-1 = 0), so (t=\frac{1}{2}).

Step3: Evaluate the function at critical - points and endpoints.

When (t=\frac{1}{2}), (y = (\frac{1}{2})^{2}-\frac{1}{2}=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}). When (t=-1), (y=(-1)^{2}-(-1)=1 + 1=2). When (t = 1), (y=1^{2}-1=0).

Step4: Recall the relationship between (t) and (x).

We were originally working with (x) values. But we found the extreme values of the function in terms of (t=\cos x). The maximum value of (y) is (2) and the minimum value is (-\frac{1}{4}).

Answer:

Absolute Maximum: (f(x)=2) Absolute Minimum: (f(x)=-\frac{1}{4})