Question

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

Hint:

You can substitute \( u = \cos x \) to treat this as a quadratic optimization problem \( g(u) = 1 + u - u^2 \) on \( u \in [-1, 1] \).
The vertex of a parabola \( -u^2 + u + 1 \) occurs at \( u = -b/2a = -1/(-2) = 1/2 \).

Answer 1

Concept:
• To find the absolute maximum of a function on a closed interval, we evaluate the function at its critical points (where \( f'(x) = 0 \)) and at the endpoints of the interval.
• Trigonometric identity: \( \sin^2 x = 1 - \cos^2 x \).

Step 1: Simplify the function expression
Rewrite the function in terms of a single trigonometric ratio (cosine):
\[ f(x) = \cos x + (1 - \cos^2 x) \]
\[ f(x) = 1 + \cos x - \cos^2 x \]

Step 2: Find critical points by differentiating
Differentiate \( f(x) \) with respect to \( x \):
\[ f'(x) = 0 - \sin x - 2\cos x (-\sin x) \]
\[ f'(x) = -\sin x + 2\sin x \cos x \]
\[ f'(x) = \sin x (2\cos x - 1) \]
Set \( f'(x) = 0 \) to find stationary points:
1) \( \sin x = 0 \implies x = 0, \pi \) (These are the endpoints of our interval).
2) \( 2\cos x - 1 = 0 \implies \cos x = 1/2 \implies x = \pi/3 \).

Step 3: Evaluate the function at critical points and endpoints
The candidate points are \( x = 0, \pi/3, \pi \).
At \( x = 0 \):
\[ f(0) = 1 + \cos 0 - \cos^2 0 = 1 + 1 - 1^2 = 1 \]
At \( x = \pi/3 \):
\[ f(\pi/3) = 1 + \cos(\pi/3) - \cos^2(\pi/3) = 1 + 1/2 - (1/2)^2 \]
\[ f(\pi/3) = 1 + 0.5 - 0.25 = 1.25 = 5/4 \]
At \( x = \pi \):
\[ f(\pi) = 1 + \cos\pi - \cos^2\pi = 1 + (-1) - (-1)^2 \]
\[ f(\pi) = 1 - 1 - 1 = -1 \]

Step 4: Identify the absolute maximum
Comparing the values: \( f(0) = 1 \), \( f(\pi/3) = 1.25 \), and \( f(\pi) = -1 \).
The absolute maximum value is \( 5/4 \) or \( 1.25 \).