Maxima and minima

Walk along the graph of almost any smooth function and you pass over hilltops and through valleys — points where the curve stops rising and starts falling, or the reverse. At every such point the tangent line is momentarily flat: the slope $f'(x)$ is zero. That single geometric fact is the seed of an entire chapter, Applications of Derivatives, and this article develops it into two working tools — the first-derivative test and the second-derivative test — for finding and classifying these local high and low points, called local maxima and local minima. The payoff is not just graph-sketching. The same machinery turns real optimization questions — the cheapest box, the shortest fence, the most profitable price — into a short calculus recipe: write the quantity as a function of one variable, differentiate, find where the slope vanishes, and classify. This article carries that recipe through both a pure algebraic example and a real fencing problem, and flags the traps that cost the most marks in this topic.

The core insight

A differentiable function can have a local maximum or minimum only where its derivative is zero, and the sign of the second derivative there — negative for a peak, positive for a valley, zero for "test fails, check the first derivative instead" — tells you which.

The core idea

A function $f$ has a local maximum at $x = c$ if $f(c) \geq f(x)$ for all $x$ near $c$ , and a local minimum at $x = c$ if $f(c) \leq f(x)$ for all $x$ near $c$ . Together these are called local extrema, and the value $f(c)$ is the corresponding extreme value. Note the word "local": the comparison is only with nearby points, not with the function's behaviour far away.

Critical points come first. If $f$ is differentiable at $c$ and has a local extremum there, then necessarily $f'(c) = 0$ — this condition is necessary, not sufficient, since a horizontal tangent can also occur at a point that is neither a peak nor a valley. Points where $f'(x) = 0$ are called stationary points; together with any points where $f'$ fails to exist, they form the critical points, which are the only candidates worth testing.

f'(c) = 0 \quad \text{(necessary condition for a smooth local extremum at } x=c\text{)}.

where: $f'(c)$ is the derivative of $f$ evaluated at the candidate point $c$ ; the equation says the tangent is horizontal there, nothing more.

First-derivative test. Once a critical point $c$ is found, check how the sign of $f'(x)$ changes as $x$ crosses $c$ . If $f'(x)$ switches from positive (rising) to negative (falling), $f$ peaks at $c$ — a local maximum. If it switches from negative to positive, $f$ bottoms out at $c$ — a local minimum. If the sign does not change at all, $c$ is neither; it is a point of inflection with a momentarily flat tangent (the classic example is $f(x) = x^3$ at $x = 0$ ).

Second-derivative test. This is usually faster when $f''$ is easy to compute. At a stationary point $c$ (so $f'(c) = 0$ ):

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

where: $f''(c)$ is the second derivative — the rate at which the slope itself is changing. A negative $f''(c)$ means the curve is bending downward (concave down) near $c$ , consistent with a peak; a positive $f''(c)$ means it bends upward (concave up), consistent with a valley. When $f''(c) = 0$ the curve's bending is too mild to decide, and you must fall back on the first-derivative test. For applied optimization problems the strategy is always the same three moves: express the quantity to be optimized as a function of a single variable using the given constraint, differentiate and set the derivative to zero to find critical points, then classify with either test.

Figure

A cubic curve showing a local maximum on the left and a local minimum on the right, each with a horizontal dashed tangent line marking the zero slope

Derivation AClassifying the critical points of a cubic using both tests

The goal is to find and classify every local extremum of the cubic $f(x) = x^3 - 3x^2 - 9x + 5$ , showing step by step how the necessary condition $f'(x)=0$ locates the candidates and how the second-derivative test decides which candidate is a peak and which is a valley, ending with both the location and the actual extreme value at each point.

Step 1. Differentiate $f$ once to get the slope function: $f'(x) = 3x^2 - 6x - 9$ . This can be factored as $f'(x) = 3(x^2 - 2x - 3) = 3(x-3)(x+1)$ , which makes the zeros easy to read off directly.

Step 2. Set $f'(x) = 0$ to find the critical points: $3(x-3)(x+1) = 0$ gives $x = 3$ and $x = -1$ . These are the only two places the tangent to $f$ is horizontal, so they are the only candidates for a local extremum.

Step 3. Differentiate again to get $f''(x) = 6x - 6$ , and evaluate it at each candidate to classify. At $x = -1$ : $f''(-1) = 6(-1) - 6 = -12 < 0$ , so $f$ has a local maximum at $x = -1$ . At $x = 3$ : $f''(3) = 6(3) - 6 = 12 > 0$ , so $f$ has a local minimum at $x = 3$ .

Step 4. Report the actual extreme values by substituting back into $f$ , not just the locations. $f(-1) = (-1)^3 - 3(-1)^2 - 9(-1) + 5 = -1 - 3 + 9 + 5 = 10$ , and $f(3) = 27 - 27 - 27 + 5 = -22$ . So the local maximum value is $10$ at $x=-1$ , and the local minimum value is $-22$ at $x=3$ — matching the shape of the graph above, one peak followed by one valley as $x$ increases.

Worked example

A farmer wants to fence a rectangular field of area $200\ \text{m}^2$ using the minimum possible length of fencing. Find the dimensions that minimize the perimeter.

Step 1. Set up one variable using the constraint. Let the sides be $x$ and $y$ metres, so $xy = 200$ , which gives $y = \dfrac{200}{x}$ for $x > 0$ . The perimeter to minimize is $P(x) = 2x + 2y = 2x + \dfrac{400}{x}$ , now written purely in terms of $x$ .

Step 2. Differentiate and find the critical point. $P'(x) = 2 - \dfrac{400}{x^2}$ . Setting $P'(x) = 0$ gives $2 = \dfrac{400}{x^2}$ , so $x^2 = 200$ and $x = \sqrt{200} = 10\sqrt{2}$ metres (taking the positive root since $x$ is a length).

Step 3. Confirm it is a minimum using the second derivative. $P''(x) = \dfrac{800}{x^3}$ , and since $x = 10\sqrt{2} > 0$ , $P''(x) > 0$ — confirming a local minimum, not a maximum.

Step 4. Find the matching $y$ and state the answer. $y = \dfrac{200}{10\sqrt{2}} = 10\sqrt{2}$ , so $x = y = 10\sqrt{2}\ \text{m} \approx 14.14\ \text{m}$ : the field is a square of side $10\sqrt{2}$ m, giving the minimum perimeter $P = 4(10\sqrt2) = 40\sqrt2 \approx 56.6$ m.

Common traps

Treating every stationary point as an extremum. $f'(c) = 0$ is necessary but not sufficient — a point of inflection with a horizontal tangent (like $f(x)=x^3$ at $x=0$ ) is neither a maximum nor a minimum.
Confusing the extreme point with the extreme value. The critical point $x = c$ tells you where the extremum occurs; you must substitute back into $f$ to report the actual maximum or minimum value.
Forgetting the inconclusive case. When $f''(c) = 0$ , the second-derivative test decides nothing — switch to the first-derivative sign-change test instead of guessing.
Not reducing an optimization problem to one variable. In applied problems with a constraint, eliminate one variable first; differentiating an expression that still has two independent variables is a common setup error.

Ask the tutor: Want to try classifying a trickier critical point, or work through another applied optimization problem such as minimizing the cost of an open box? Ask the AI tutor to walk you through it.

Why is f prime of c equals zero not enough to guarantee an extremum?

Because a horizontal tangent can also occur at a point where the curve is simply pausing on its way up or down, not turning around. The standard example is $f(x) = x^3$ at $x = 0$ : here $f'(0) = 0$ , but for $x$ slightly negative $f(x) < 0$ and for $x$ slightly positive $f(x) > 0$ , so the function never actually stops increasing near $x=0$ — it is a point of inflection, not an extremum. That is why $f'(c)=0$ only narrows down the candidates; you still need the first- or second-derivative test to confirm what kind of point it actually is.

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

Use the second-derivative test first because it is usually quicker — just compute $f''(c)$ and check its sign. Switch to the first-derivative test whenever $f''(c) = 0$ (the second-derivative test is inconclusive there) or whenever $f''$ is awkward or impossible to compute cleanly. The first-derivative test is also more fundamental: it works even at points where $f'$ exists but $f''$ does not, since it only needs the sign of $f'$ just before and just after the critical point.

In word problems how do I know whether I am looking for a maximum or a minimum?

Read the question's wording carefully: "minimum fencing," "least cost," or "shortest length" call for a local minimum, while "maximum area," "greatest volume," or "highest profit" call for a local maximum. Once you differentiate and find the critical point, do not assume which one it is — always confirm with the second-derivative test (or first-derivative test), as in the fencing example where $P''(x) > 0$ genuinely confirmed a minimum rather than just being assumed from the wording.

Maxima and minima

The core idea

Worked example

Common traps

Before this

Where this leads