Maxima and minima

Every hill and every dip on a smooth curve gives itself away the same way: right at the top of a hill or the bottom of a dip, the tangent line goes momentarily flat. That single geometric fact — the tangent flattening out — is the seed of an entire method for finding where a function is largest or smallest, and it is the mathematical engine behind every "best possible" question you will ever meet: the cheapest box, the fastest route, the maximum profit. This article builds that method from the ground up. It defines stationary points, local versus absolute extrema, and the two standard tests (first-derivative and second-derivative) used to classify a critical point, then walks through fully worked examples on both an open-ended cubic and a closed-interval quadratic, and closes with the traps that cost the most marks in "Applications of Derivatives."

The core insight

At a smooth local maximum or minimum the derivative vanishes, $f'(c)=0$ , and the sign of $f'$ around $c$ (or the sign of $f''(c)$ ) tells you whether that flat point is a peak, a valley, or neither.

The core idea

Stationary (critical) points. A point $x=c$ in the domain of $f$ is called a stationary point (or critical point) if $f'(c)=0$ . This syllabus restricts attention to differentiable functions, so every critical point here is a genuine flat-tangent point — no corners or vertical tangents to worry about.

Local versus absolute extrema. $f$ has a local maximum at $x=c$ if $f(c)\ge f(x)$ for every $x$ in some small neighbourhood of $c$ — it only has to beat its immediate neighbours, not the whole function. A local minimum is defined the same way with $\le$ . An absolute (global) maximum or minimum, by contrast, is the largest or smallest value of $f$ over its entire domain, or over a specified closed interval $[a,b]$ . Absolute extrema are found by comparing the values of $f$ at all interior critical points against the values of $f$ at the two endpoints $a$ and $b$ — a local extremum is only a candidate, and the endpoints must always be checked too.

The first-derivative test. At a critical point $c$ , look at the sign of $f'(x)$ just to the left and just to the right of $c$ :

f'(x)>0 \text{ for } x<c \text{ and } f'(x)<0 \text{ for } x>c \implies \text{local maximum at } c

where: the sign change from positive to negative means $f$ was rising just before $c$ and falling just after — exactly the shape of a peak. The mirror case, negative-to-positive, gives a local minimum, and no sign change at all (positive on both sides, or negative on both sides) means $c$ is neither a maximum nor a minimum — typically a point of inflection with a momentarily flat tangent.

The second-derivative test. This is the faster shortcut once a critical point is found:

f''(c) < 0 \implies \text{local maximum at } c, \qquad f''(c) > 0 \implies \text{local minimum at } c

where: $f''(c)$ measures concavity at $c$ — negative concavity means the curve bends downward like a dome (a peak), positive concavity means it bends upward like a bowl (a valley). If $f''(c)=0$ the test is silent and gives no information either way; you must fall back on the first-derivative test at that point. The two tests are not competitors so much as a fast-then-slow pair: try the quick single evaluation $f''(c)$ first, and only do the more laborious sign-tracking of $f'$ on both sides of $c$ if $f''(c)$ turns out to be zero.

Figure

A curve with a hill shaped peak on the left and a valley shaped dip on the right, flat tangent lines drawn at both the peak and the dip, and arrows showing the curve rising then falling around the peak and falling then rising around the dip

Derivation AWhy a smooth local extremum must have f'(c)=0

The goal of this derivation is to establish the necessary condition every student uses without always seeing why it is true: if a differentiable function $f$ has a local maximum or local minimum at an interior point $x=c$ , then the derivative $f'(c)$ must equal exactly zero, so that hunting for extrema always starts by solving $f'(x)=0$ . The argument uses nothing beyond the definition of the derivative as a limit and the definition of local maximum.

Step 1. Suppose $f$ has a local maximum at $x=c$ . By definition, there is some small neighbourhood of $c$ in which $f(c)\ge f(x)$ for every $x$ nearby, so $f(x)-f(c)\le 0$ throughout that neighbourhood.

Step 2. Consider the derivative from the right, using $x=c+h$ with $h>0$ small. The difference quotient is $\dfrac{f(c+h)-f(c)}{h}$ . The numerator is $\le 0$ (from Step 1) and $h>0$ , so the whole quotient is $\le 0$ . Taking the limit as $h\to 0^+$ , the right-hand derivative satisfies $f'(c)\le 0$ .

Step 3. Now consider the derivative from the left, using $x=c-h$ with $h>0$ small. The difference quotient is $\dfrac{f(c-h)-f(c)}{-h}$ . The numerator $f(c-h)-f(c)$ is again $\le 0$ , but now it is divided by $-h<0$ , so the quotient is $\ge 0$ . Taking the limit as $h\to 0^-$ , the left-hand derivative satisfies $f'(c)\ge 0$ .

Step 4. Since $f$ is given to be differentiable at $c$ , the left-hand and right-hand derivatives must be equal to the single value $f'(c)$ . Step 2 forces $f'(c)\le 0$ and Step 3 forces $f'(c)\ge 0$ ; the only number satisfying both is $f'(c)=0$ . The identical argument with the inequality reversed proves the same conclusion for a local minimum. This is exactly why every extremum search on this syllabus begins with solving $f'(x)=0$ — it is a necessary condition, though Worked example below shows it is not by itself sufficient, since not every critical point turns out to be an extremum.

Worked example

Find the local maximum and minimum values of $f(x) = x^3-3x^2-9x+5$ , and also determine the absolute maximum and minimum of the same expression $g(x)=x^2-4x+3$ on the closed interval $[0,5]$ .

Step 1 — Differentiate and locate critical points of $f$ . $f'(x) = 3x^2-6x-9 = 3(x-3)(x+1)$ , so $f'(x)=0$ gives $x=3$ or $x=-1$ . These are the only candidates for a local extremum, by the necessary condition proved above.

Step 2 — Classify each critical point with the second-derivative test. $f''(x)=6x-6$ . At $x=-1$ : $f''(-1)=-6-6=-12<0$ , so $f$ has a local maximum there. At $x=3$ : $f''(3)=18-6=12>0$ , so $f$ has a local minimum there.

Step 3 — Compute the actual extreme values. $f(-1) = (-1)^3-3(-1)^2-9(-1)+5 = -1-3+9+5 = 10$ , and $f(3) = 27-27-27+5=-22$ . So $f$ has a local maximum value of $\boxed{10}$ at $x=-1$ and a local minimum value of $\boxed{-22}$ at $x=3$ .

Step 4 — Now find absolute extrema of $g(x)=x^2-4x+3$ on $[0,5]$ . First find interior critical points: $g'(x)=2x-4=0 \implies x=2$ , which lies inside $[0,5]$ . Evaluate $g$ at this critical point and at both endpoints: $g(0)=3$ , $g(2)=4-8+3=-1$ , $g(5)=25-20+3=8$ .

Step 5 — Compare all three candidate values. The set of values is $\{3,-1,8\}$ , so the absolute maximum is $\boxed{8}$ at $x=5$ and the absolute minimum is $\boxed{-1}$ at $x=2$ . Notice the absolute maximum sits at an endpoint, not at the interior stationary point — the reason absolute-extrema questions always require checking endpoints, never just critical points.

Common traps

Assuming every critical point is automatically a local maximum or minimum — some are points of inflection with a horizontal tangent, where $f'$ does not change sign at all.
Forgetting to check the interval endpoints when hunting for absolute extrema on a closed interval, as Worked example Step 5 shows directly.
Misapplying the second-derivative test when $f''(c)=0$ — this is inconclusive, not automatically "neither," and demands falling back on the first-derivative test.
Confusing "local" with "absolute": a local maximum only has to beat its immediate neighbours, and need not be the largest value of the function anywhere else.

Ask the tutor: stuck on whether a stationary point is a maximum, a minimum, or neither — ask the AI tutor to walk the sign-change analysis with you on your own function.

If f'(c)=0, does that always mean x=c is a local max or min?

No. $f'(c)=0$ is only a necessary condition, established by the derivation above, not a guarantee. Classic example: $f(x)=x^3$ has $f'(0)=0$ , but $f$ is increasing on both sides of $x=0$ , so it is neither a local maximum nor a local minimum there — it is a point of inflection with a momentarily flat tangent. Always follow up a stationary point with the first- or second-derivative test before declaring it an extremum.

When should I use the first-derivative test instead of the second-derivative test?

Use the second-derivative test first, since it is usually a single quick computation of $f''(c)$ . Switch to the first-derivative test only when $f''(c)=0$ , because the second-derivative test genuinely gives no information in that case — it is not evidence of "neither max nor min," it simply fails to decide. The first-derivative test always works, even when the second-derivative test is silent, because it directly tracks whether $f$ is increasing or decreasing on each side of $c$ .

Why do absolute extrema need endpoint checks but local extrema don't?

A local extremum only compares $f(c)$ against its immediate neighbours, so it is entirely a statement about the interior behaviour of the curve near $c$ , and endpoints of an interval are irrelevant to that local comparison. An absolute extremum, however, asks for the single largest or smallest value across the whole interval, including its boundary — and a closed interval's endpoints are legitimate points in the domain that a stationary-point search never visits, since $g'(x)$ need not be zero there. Worked example Step 5 shows exactly this: the absolute maximum of $g$ occurred at the endpoint $x=5$ , not at the interior critical point $x=2$ .

Maxima and minima

The core idea

Worked example

Common traps

Before this

Where this leads