Q.Find the absolute maximum and minimum values of the function $f$ given by $f(x) = \cos^2 x + \sin x, x \in [0, \pi]$

Idea: On a closed interval, the absolute extrema occur at critical points or endpoints — evaluate $f$ at all of them and compare.

$f(x)=\cos^2 x+\sin x$ on $[0,\pi]$.

Set $f'(x)=0$: either $\cos x=0\Rightarrow x=\tfrac{\pi}{2}$, or $\sin x=\tfrac12\Rightarrow x=\tfrac{\pi}{6},\tfrac{5\pi}{6}$.

Evaluate at critical points and endpoints:

• $f(0)=1+0=1$
• $f\!\left(\tfrac{\pi}{6}\right)=\tfrac34+\tfrac12=\tfrac54$
• $f\!\left(\tfrac{\pi}{2}\right)=0+1=1$ …

The method

$f$ is continuous on the closed interval $[0,\pi]$, so by the closed-interval method the absolute maximum and minimum each occur either at a critical point (where $f'=0$) or at an endpoint. We find all such points and compare the values of $f$.

Step 1 — A helpful simplification

Use $\cos^2x=1-\sin^2x$:

Let $t=\sin x$. On $[0,\pi]$, $\sin x$ runs from $0$ up to $1$ and back to $0$, so $t\in[0,1]$, and

This is a downward parabola with vertex at $t=-\dfrac{b}{2a}=\dfrac12$, which lies in $[0,1]$.

Step 2 — Critical points via $f'$

Equivalently, differentiate directly:

Set $f'(x)=0$:

• $\cos x=0\Rightarrow x=\dfrac{\pi}{2}$;
• $1-2\sin x=0\Rightarrow \sin x=\dfrac12\Rightarrow x=\dfrac{\pi}{6},\ \dfrac{5\pi}{6}$ (both in $[0,\pi]$).

Step 3 — Evaluate at critical points and endpoints

$f\!\left(\tfrac{\pi}{2}\right)=0+1=1,$ …

Method: Finding Absolute Extrema on a Closed Interval (Closed Interval Method)

This method finds the absolute (global) maximum and minimum values of a continuous function on a closed interval $[a,b]$ — the standard technique for any "find the absolute maximum and minimum" question with a stated closed domain.

Steps

Step 1: Confirm $f$ is continuous on the closed interval given

The Closed Interval Method only guarantees the absolute extrema exist (and are attained) when $f$ is continuous on $[a,b]$ — check this before proceeding.

Step 2: Simplify $f$ if a substitution reveals a simpler structure

If the function is built from a single repeated expression (like $\sin x$ appearing as both $\sin x$ and, via $\cos^2x=1-\sin^2x$, hidden inside $\cos^2x$), substituting $t=$ that expression can turn a trig function into a simple polynomial in $t$ — useful for understanding the shape of the graph, though the critical points must still be found in terms of the original variable.

Step 3: Differentiate $f$ and find all critical points inside $(a,b)$

Solve $f'(x)=0$ for every solution that lies strictly inside the interval (discard any that fall outside $[a,b]$), and note any points where $f'$ fails to exist. …

Common Mistakes

Mistake 1: Forgetting to check the endpoints of the interval

Why it's wrong: the absolute extrema of a function on a closed interval can occur at the endpoints even when they aren't critical points — evaluating only the critical points and skipping $f(a)$ and $f(b)$ can miss the true maximum or minimum entirely. Correct approach: always compute $f$ at both endpoints as a mandatory part of the comparison list, never as an afterthought.

Mistake 2: Missing a second solution when solving $\sin x = c$ or $\cos x = c$ on the given interval

Why it's wrong: an equation like $\sin x=\tfrac12$ has two solutions in $[0,\pi]$ (here $x=\pi/6$ and $x=5\pi/6$), since $\sin x$ is not one-to-one over the whole interval — finding only one root under-counts the critical points and can silently drop a location where the extreme value repeats. Correct approach: sketch or recall the shape of the relevant trig function over the given interval and check for every solution, not just the first one found. …