Question

Find the absolute maximum value of $f(x)=\cos x+\sin^2 x$, where $x\in[0,\pi]$.

Hint:

Since the interval is closed and continuous, the function must attain an absolute maximum and minimum.

Answer 1

Step 1: Write the given function.
The function is \[ f(x)=\cos x+\sin^2 x \] defined on the closed interval \[ x\in[0,\pi] \] Since the interval is closed and continuous, the function must attain an absolute maximum and minimum.
Step 2: Differentiate the function.
Differentiate with respect to \(x\): \[ f'(x)=-\sin x+2\sin x\cos x \] Factor out \(\sin x\): \[ f'(x)=\sin x(2\cos x-1) \] Step 3: Find critical points.
Set \[ f'(x)=0 \] Thus \[ \sin x(2\cos x-1)=0 \] This gives \[ \sin x=0 \] or \[ 2\cos x-1=0 \] First case: \[ \sin x=0 \] \[ x=0,\pi \] Second case: \[ 2\cos x-1=0 \] \[ \cos x=\frac12 \] \[ x=\frac{\pi}{3} \] Step 4: Evaluate the function at critical points and endpoints.
Compute \(f(x)\): For \(x=0\): \[ f(0)=\cos0+\sin^20 \] \[ =1 \] For \(x=\pi\): \[ f(\pi)=\cos\pi+\sin^2\pi \] \[ =-1 \] For \(x=\frac{\pi}{3}\): \[ f\left(\frac{\pi}{3}\right)=\cos\frac{\pi}{3}+\sin^2\frac{\pi}{3} \] \[ =\frac12+\left(\frac{\sqrt3}{2}\right)^2 \] \[ =\frac12+\frac34 \] \[ =\frac54 \] Step 5: Determine the absolute maximum.
The values obtained are \[ 1,\;-1,\;\frac54 \] The largest value is \[ \frac54 \] Final Answer: \[ \boxed{\frac54} \]