Q.If $\phi(x) = f(x) + f(1-x)$, $f''(x) = 0$ for $0 \le x \le 1$, then is $x = \frac{1}{2}$ a point of maxima or minima of $\phi(x)$?

Critical Points Analysis: Where Functions Change Direction

Hiking a mountain range, you reach peaks (highest spot around), valleys (bottoms), and flat stretches where the ground doesn't slope. These special locations — peaks, valleys, and flat spots — are critical points.

The Intuition

A function's graph is like that trail. At most points it is rising (positive slope) or falling (negative slope). At a critical point something changes: the slope becomes zero, or the slope doesn't exist (a sharp corner).

Throw a ball straight up: at the very top of its arc it stops for an instant before falling. Its velocity — the rate of change of height — is zero at that moment. That's a critical point.

The Precise Definition

Why Two Conditions?

Derivative equals zero catches the "flat" spots — peaks, valleys, horizontal plateaus — where the tangent line is horizontal.

Derivative does not exist catches sharp corners (like the tip of $|x|$ at $x=0$), vertical tangents, and cusps. Even without a zero slope, these can be peaks or valleys.

How to Find Critical Points

1. Find the derivative $f'(x)$.
2. Solve $f'(x) = 0$ — these are candidates.
3. Check where $f'(x)$ does not exist — but only if $f(x)$ exists there (the point must be in the domain).
4. Collect all such $x$-values.

Example 1: A Simple Polynomial

Let $f(x) = x^3 - 3x^2 + 1$.

$f'(x) = 3x^2 - 6x = 3x(x - 2)$.

$f'(x) = 0 \implies x = 0$ or $x = 2$. Since $f'$ exists everywhere, the critical points are $x = 0$ and $x = 2$.

Example 2: A Function with a Corner

Let $f(x) = |x|$. Here $f'(x)$ does not exist at $x = 0$ (left derivative $-1$, right derivative $+1$), and $f'(x) = 0$ has no solutions. So the only critical point is $x = 0$.

What Critical Points Tell Us …