Critical Points and Extrema

Where is a function highest? Where is it lowest? These questions come up everywhere: the top of a ball’s flight, the lowest cost, the largest volume. This page sets up the vocabulary (absolute and relative extrema), tells you when a highest and lowest point are guaranteed to exist, and shows where to look for them: at critical points.

Key ideas

Absolute and relative extrema

“Extrema” is the plural of “extremum”: a maximum or a minimum.

• $f$ has an absolute maximum (or global maximum) at $x = c$ if $f(c) \ge f(x)$ for every $x$ in the domain (or interval) you’re looking at. The value $f(c)$ is the absolute maximum value.
• $f$ has a relative maximum (or local maximum) at $x = c$ if $f(c) \ge f(x)$ for all $x$ near $c$ (on both sides of $c$): it’s the top of a hill, even if there’s a taller hill somewhere else.

Minimums are defined the same way with $\le$. An absolute extremum can also be a relative one, and it can happen at an endpoint of an interval.

On $[0.5, 4.2]$, the top of the hill at $x = 1$ is only a relative maximum: the right endpoint is higher.

Textbooks differ on whether an endpoint can count as a relative extremum. To be safe, look for relative extrema at interior points, and always check endpoints when you want absolute extrema.

The Extreme Value Theorem

Extreme Value Theorem (EVT). If $f$ is continuous on a closed interval $[a, b]$, then $f$ has both an absolute maximum and an absolute minimum on $[a, b]$.

Both conditions matter:

• On an open interval, there may be no maximum. $f(x) = x^2$ on $(0, 2)$ gets close to $4$ but never reaches it, because $x = 2$ isn’t included.
• With a discontinuity, there may be no maximum. $f(x) = \dfrac{1}{x - 1}$ on $[0, 2]$ shoots up to $\infty$ near $x = 1$.

Like the Mean Value Theorem, the EVT guarantees that the extrema exist; it doesn’t tell you where they are.

Critical points

A critical point (or critical number) of $f$ is a number $c$ in the domain of $f$ where

$$f'(c) = 0 \qquad \text{or} \qquad f'(c) \text{ does not exist}$$

• $f'(c) = 0$: a horizontal tangent, like the top of a smooth hill.
• $f'(c)$ undefined: a corner, a cusp, or a vertical tangent.

If $f$ isn’t defined at $c$, then $c$ is not a critical point, even if $f'$ is undefined there.

Why critical points matter

If $f$ has a relative extremum at an interior point $x = c$, then $c$ must be a critical point. (At the top of a smooth hill the tangent is flat; otherwise the hilltop is a sharp point where $f'$ doesn’t exist.)

The reverse is not true: a critical point doesn’t have to be an extremum. For $f(x) = x^3$, $f'(0) = 0$, but the graph keeps rising through $x = 0$. Critical points are candidates. The first derivative test and the candidates test decide which candidates win.

Worked examples

Example 1: Critical points of a polynomial

Find the critical points of $f(x) = 2x^3 - 3x^2 - 12x + 4$.

Solution. A polynomial’s derivative exists everywhere, so the only critical points are where $f'(x) = 0$:

$$f'(x) = 6x^2 - 6x - 12 = 6(x^2 - x - 2) = 6(x - 2)(x + 1)$$

The critical points are $x = -1$ and $x = 2$.

Example 2: Where the derivative doesn’t exist

Find the critical points of $f(x) = x^{2/3}(x - 5)$.

Solution. Expand first, so you can use the power rule:

$$f(x) = x^{5/3} - 5x^{2/3}$$ $$\begin{aligned} f'(x) &= \frac{5}{3}x^{2/3} - \frac{10}{3}x^{-1/3} \\ &= \frac{5}{3}x^{-1/3}(x - 2) && \text{factor out } \tfrac{5}{3}x^{-1/3} \\ &= \frac{5(x - 2)}{3x^{1/3}} \end{aligned}$$

• $f'(x) = 0$ when the numerator is $0$: $x = 2$.
• $f'(x)$ is undefined when the denominator is $0$: $x = 0$. And $f(0) = 0$ is defined, so $x = 0$ is in the domain.

The critical points are $x = 0$ and $x = 2$. (The graph has a cusp at $x = 0$.)

Example 3: Trig critical points (radians)

Find the critical points of $f(x) = \sin x + \cos x$ on $[0, 2\pi]$. Calculus always uses radians: $\pi$ radians is $180^\circ$.

Solution. $f'(x) = \cos x - \sin x$, which exists everywhere. Set it equal to $0$:

$$\cos x = \sin x \quad\Rightarrow\quad \tan x = 1$$

(Dividing by $\cos x$ is fine: if $\cos x = 0$, then $\sin x = \pm 1$, so they can’t be equal.) On $[0, 2\pi]$, $\tan x = 1$ at $x = \dfrac{\pi}{4}$ and $x = \dfrac{5\pi}{4}$.

Example 4: Is an absolute maximum guaranteed?

For each function, does the EVT guarantee an absolute maximum and minimum on the interval?

• (a) $f(x) = x^3 - 4x$ on $[-1, 3]$
• (b) $g(x) = \dfrac{1}{x}$ on $[-1, 1]$
• (c) $h(x) = \sqrt{x}$ on $(0, 4)$

Solution.

(a) Yes. $f$ is a polynomial, so it’s continuous on the closed interval $[-1, 3]$.

(b) No. $g$ is not continuous at $x = 0$, which is in the interval. (In fact $g$ has no absolute max or min there: it heads to $\pm\infty$ near $0$.)

(c) No. The interval is open. In fact $h$ has neither: its values get close to $0$ and $2$ but never reach them, since $x = 0$ and $x = 4$ are left out.

Common mistakes

Forgetting the points where f′ is undefined. After solving $f'(x) = 0$, look at the denominator of $f'$ too. In Example 2, missing $x = 0$ would miss the cusp.

Calling a point a critical point when it isn’t in the domain. For $f(x) = \dfrac{1}{x}$, $f'$ is undefined at $x = 0$, but so is $f$. So $x = 0$ is not a critical point.

Assuming every critical point is a maximum or minimum. $f(x) = x^3$ has a critical point at $0$ and no extremum. You need a test to decide.

Using the EVT without checking both conditions. Write ”$f$ is continuous on the closed interval $[a, b]$” before using the theorem. Open intervals and discontinuities break it.

Giving the x-value when the question asks for the value. “The absolute maximum value” is $f(c)$, a $y$-value. “Where” or “at what $x$” asks for $c$. Read the question carefully.

Practice

1. (Warm-up) Find the critical point of $f(x) = x^2 - 8x + 3$.

Solution

$f'(x) = 2x - 8 = 0$ gives $x = 4$.

2. (Warm-up) Find the critical points of $f(x) = x^3 - 12x$.

Solution

$$f'(x) = 3x^2 - 12 = 3(x - 2)(x + 2)$$

The critical points are $x = -2$ and $x = 2$.

3. (Warm-up) Does the EVT guarantee that $f(x) = \dfrac{1}{x - 3}$ has an absolute maximum on $[0, 2]$? On $[2, 4]$?

Solution

On $[0, 2]$: yes. The only discontinuity is at $x = 3$, which is outside the interval, so $f$ is continuous on the closed interval $[0, 2]$.

On $[2, 4]$: no. $x = 3$ is inside the interval, so $f$ is not continuous there. (It actually has no maximum: it goes to $\infty$ as $x \to 3^+$.)

4. (Core) Find the critical points of $f(x) = x^4 - 4x^3$.

Solution

$$f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)$$

The critical points are $x = 0$ and $x = 3$.

5. (Core) Find the critical points of $f(x) = (x^2 - 4)^{2/3}$.

Solution

By the chain rule:

$$f'(x) = \frac{2}{3}(x^2 - 4)^{-1/3} \cdot 2x = \frac{4x}{3\sqrt[3]{x^2 - 4}}$$

• $f'(x) = 0$ when $x = 0$.
• $f'(x)$ is undefined when $x^2 - 4 = 0$, so $x = \pm 2$. Since $f(\pm 2) = 0$ is defined, these are in the domain.

The critical points are $x = -2$, $x = 0$, and $x = 2$.

6. (Core) Find the critical points of $f(x) = x - 2\sin x$ on $[0, 2\pi]$ (radians).

Solution

$f'(x) = 1 - 2\cos x$, which exists everywhere. Set it to $0$:

$$\cos x = \frac{1}{2} \quad\Rightarrow\quad x = \frac{\pi}{3} \ \text{ or } \ x = \frac{5\pi}{3}$$

7. (Core) Find the critical points of $f(x) = \dfrac{x}{x^2 + 9}$.

Solution

By the quotient rule:

$$f'(x) = \frac{(x^2 + 9)(1) - x(2x)}{(x^2 + 9)^2} = \frac{9 - x^2}{(x^2 + 9)^2}$$

The denominator is never $0$, so $f'$ always exists. $f'(x) = 0$ when $9 - x^2 = 0$: $x = -3$ and $x = 3$.

8. (Challenge) Show that $f(x) = \dfrac{x - 1}{x + 2}$ has no critical points, even though $f'$ is undefined at $x = -2$.

Solution

$$f'(x) = \frac{(x + 2)(1) - (x - 1)(1)}{(x + 2)^2} = \frac{3}{(x + 2)^2}$$

The numerator is $3$, so $f'(x)$ is never $0$. $f'$ is undefined only at $x = -2$, but $f(-2)$ is undefined too (division by zero), so $-2$ is not in the domain and isn’t a critical point. So $f$ has no critical points.

9. (Challenge) Find constants $a$ and $b$ so that $f(x) = x^3 + ax^2 + bx$ has critical points at $x = -1$ and $x = 3$.

Solution

$f'(x) = 3x^2 + 2ax + b$. We need $f'(-1) = 0$ and $f'(3) = 0$, so $f'$ must be $3(x + 1)(x - 3)$ (the leading coefficient is $3$):

$$3(x + 1)(x - 3) = 3x^2 - 6x - 9$$

Matching coefficients: $2a = -6$ and $b = -9$. So $a = -3$ and $b = -9$.

Check: $f'(x) = 3x^2 - 6x - 9$ gives $f'(-1) = 3 + 6 - 9 = 0$ and $f'(3) = 27 - 18 - 9 = 0$. ✓