Question

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

Hint:

Always convert trigonometric expressions into a single function type if possible when checking absolute boundaries. Alternatively, rewrite \( f(x) = \cos x + (1 - \cos^2 x) = 1 + \cos x - \cos^2 x \). Letting \( t = \cos x \) where \( t \in [-1, 1] \), you get a simple quadratic function \( g(t) = 1 + t - t^2 \), whose maximum occurs at its vertex \( t = -\frac{1}{2(-1)} = \frac{1}{2} \).

Answer 1

Concept: To determine the absolute maximum value of a continuous function \( f(x) \) on a closed bounded interval \( [a, b] \), we must find all critical points of the function within the interval and evaluate the function's value at these critical points as well as at the boundary endpoints \( x = a \) and \( x = b \).
• A point \( x = c \) is defined as a critical point if \( f'(c) = 0 \) or if \( f'(c) \) does not exist.
• The absolute maximum value is the largest value among all the computed values: \( \max \{ f(a), f(b), f(c_1), f(c_2), \ldots \} \).

Step 1: Write down the function and its domain.
The given function is: \[ f(x) = \cos x + \sin^2 x \] The given closed interval constraints are: \[ x \in [0, \pi] \]

Step 2: Find the derivative of the function with respect to \( x \).
We use the standard differentiation rules. Differentiating term by term using the chain rule on the second term: \[ f'(x) = \frac{d}{dx}(\cos x) + \frac{d}{dx}(\sin^2 x) \] \[ f'(x) = -\sin x + 2\sin x \cdot \frac{d}{dx}(\sin x) \] \[ f'(x) = -\sin x + 2\sin x \cos x \]

Step 3: Set \( f'(x) = 0 \) to determine the critical points.
\[ -\sin x + 2\sin x \cos x = 0 \] Factoring out the common term \( \sin x \): \[ \sin x (-1 + 2\cos x) = 0 \] This product equals zero if either factor is zero. This gives us two separate equations to solve:
• \( \sin x = 0 \)
• \( -1 + 2\cos x = 0 \quad \Rightarrow \quad \cos x = \frac{1}{1} = \frac{1}{2} \) Let us solve each within the domain \( x \in [0, \pi] \):
• For \( \sin x = 0 \): In the interval \( [0, \pi] \), \( \sin x = 0 \) at the boundary points \( x = 0 \) and \( x = \pi \).
• For \( \cos x = \frac{1}{2} \): In the interval \( [0, \pi] \), the cosine function is positive only in the first quadrant. Thus, \( x = \frac{\pi}{3} \). Therefore, the internal critical point is \( x = \frac{\pi}{3} \), and the boundary endpoints are \( x = 0 \) and \( x = \pi \).

Step 4: Evaluate the function \( f(x) \) at all boundary points and critical points.
Let us systematically compute the value of \( f(x) \) at these three values of \( x \): Case 1: At the left endpoint \( x = 0 \) \[ f(0) = \cos(0) + \sin^2(0) \] Since \( \cos(0) = 1 \) and \( \sin(0) = 0 \): \[ f(0) = 1 + 0^2 = 1 \] Case 2: At the internal critical point \( x = \frac{\pi}{3} \) \[ f\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) + \sin^2\left(\frac{\pi}{3}\right) \] We know that \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \) and \( \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \). Substituting these exact values: \[ f\left(\frac{\pi}{3}\right) = \frac{1}{2} + \left(\frac{\sqrt{3}}{2}\right)^2 \] \[ f\left(\frac{\pi}{3}\right) = \frac{1}{2} + \frac{3}{4} \] Taking a common denominator of 4: \[ f\left(\frac{\pi}{3}\right) = \frac{2}{4} + \frac{3}{4} = \frac{5}{4} = 1.25 \] Case 3: At the right endpoint \( x = \pi \) \[ f(\pi) = \cos(\pi) + \sin^2(\pi) \] Since \( \cos(\pi) = -1 \) and \( \sin(\pi) = 0 \): \[ f(\pi) = -1 + 0^2 = -1 \]

Step 5: Compare values to find the absolute maximum.
Let us compile all evaluated values: \[ f(0) = 1, \quad f\left(\frac{\pi}{3}\right) = \frac{5}{4}, \quad f(\pi) = -1 \] Comparing these numbers, we find: \[ -1 < 1 < \frac{5}{4} \] Hence, the absolute maximum value of the function on the given interval is exactly \( \frac{5}{4} \), which occurs at the critical point \( x = \frac{\pi}{3} \).