What Is Maxima And Minima?
Maxima and minima are the largest and smallest values of a function, taken together they are called the extrema of the function. A maximum is a point where the function is higher than its neighbours; a minimum is a point where it is lower. On the graph these are the peaks and the valleys, and at a smooth peak or valley the tangent line is flat, so the slope is zero.
That geometric fact is the whole engine of the topic. Where a smooth curve turns from rising to falling, its derivative passes through zero. So the search for maxima and minima becomes a search for the inputs where $f'(x)=0$ (or where $f'$ fails to exist), a problem calculus is built to solve.
Throughout this article we use one consistent derivative notation, $f'(x)$ for the first derivative and $f''(x)$ for the second, and a running example function:
$$f(x) = x^3 - 3x$$
What Is The Difference Between Local And Absolute Extrema?
The single most important distinction in this topic is local versus absolute.
A local (relative) maximum at $c$ means $f(c) \ge f(x)$ for all $x$ in some open interval around $c$. It is the top of one hill, even if a taller hill sits elsewhere. A local minimum is defined the same way with the inequality reversed.
An absolute (global) maximum at $c$ means $f(c) \ge f(x)$ for every $x$ in the domain. It is the single highest value the function ever reaches, and an absolute minimum is the single lowest.
A function can have many local extrema but at most one absolute maximum value and one absolute minimum value. A local extremum lives strictly inside the domain; an absolute extremum can occur inside the domain or at an endpoint.
The Extreme Value Theorem guarantees the absolute case exists under one clear hypothesis: if $f$ is continuous on a closed, bounded interval $[a, b]$, then $f$ attains both an absolute maximum and an absolute minimum somewhere on $[a, b]$. Drop continuity or drop the closed interval, and that guarantee can vanish.
How Do You Find Critical Points?
Every candidate for a local maximum or minimum is a critical point. For an interior point $c$ in the domain of $f$, we call $c$ a critical point if:
$$f'(c) = 0 \quad \text{or} \quad f'(c) \text{ does not exist.}$$
Both halves matter. Most extrema come from $f'(c)=0$ (a smooth flat spot), but some come from a sharp corner or a vertical tangent, where the derivative is undefined. The function $f(x) = |x|$ has its minimum at $x = 0$, yet $f'(0)$ does not exist. Skip the "or undefined" half of the definition and you miss that minimum entirely.
For the running function, differentiate once:
$$f'(x) = 3x^2 - 3 = 3(x-1)(x+1)$$
Set it to zero: $3(x-1)(x+1) = 0$ gives $x = -1$ and $x = 1$. This polynomial is differentiable everywhere, so there are no "undefined" cases here. The two critical points are $x = -1$ and $x = 1$. Finding them is only half the job; next you classify each one.
How Do You Use The First Derivative Test?
The first derivative test reads the sign of $f'$ on either side of a critical point, because the sign of the derivative tells you whether the function is rising or falling.
If $f'$ changes from positive to negative (rising then falling) at $c$, then $c$ is a local maximum.
If $f'$ changes from negative to positive (falling then rising) at $c$, then $c$ is a local minimum.
If $f'$ does not change sign, $c$ is neither; the curve only pauses.
Test the sign of $f'(x) = 3(x-1)(x+1)$ in the three regions the critical points create.
Table: Sign of the first derivative for $f(x)=x^3-3x$ and what each region means.
Interval | Test point | Sign of $f'(x)$ | Behaviour |
|---|---|---|---|
$x < -1$ | $x = -2$ | $3(-3)(-1) = +9$ (positive) | Rising |
$-1 < x < 1$ | $x = 0$ | $3(-1)(1) = -3$ (negative) | Falling |
$x > 1$ | $x = 2$ | $3(1)(3) = +9$ (positive) | Rising |
At $x = -1$ the sign goes positive to negative, so $x = -1$ is a local maximum. At $x = 1$ the sign goes negative to positive, so $x = 1$ is a local minimum. Reading a sign chart is the same skill as identifying increasing and decreasing intervals. Now find the actual values:
$$f(-1) = (-1)^3 - 3(-1) = -1 + 3 = 2$$ $$f(1) = (1)^3 - 3(1) = 1 - 3 = -2$$
Final answer: local maximum $2$ at $x=-1$, local minimum $-2$ at $x=1$.
How Do You Use The Second Derivative Test?
The second derivative test is often quicker because it checks concavity instead of testing signs on both sides. It applies only at a critical point where $f'(c) = 0$.
If $f'(c) = 0$ and $f''(c) < 0$, the curve is concave down at $c$, so $c$ is a local maximum.
If $f'(c) = 0$ and $f''(c) > 0$, the curve is concave up at $c$, so $c$ is a local minimum.
If $f'(c) = 0$ and $f''(c) = 0$, the test is inconclusive. It tells you nothing, and you must fall back on the first derivative test.
Concave down is the shape of a dome, concave up is the shape of a cup; the link to concave and convex shapes is exact. Differentiate the running function a second time:
$$f''(x) = 6x$$
Evaluate at each critical point:
$$f''(-1) = 6(-1) = -6 < 0 \quad \Rightarrow \quad \text{local maximum at } x=-1$$ $$f''(1) = 6(1) = 6 > 0 \quad \Rightarrow \quad \text{local minimum at } x=1$$
Both tests agree, which is the reassurance you want. The inconclusive case is real, though: for $f(x)=x^4$, $f'(0)=0$ and $f''(0)=0$, yet $x=0$ is a minimum; for $f(x)=x^3$, $f'(0)=0$ and $f''(0)=0$, yet $x=0$ is not an extremum at all. When $f''(c)=0$, the second derivative test simply steps aside.
How Do You Find Absolute Extrema On A Closed Interval?
To find the absolute maximum and minimum of a continuous function on a closed interval $[a, b]$, use the closed-interval method:
Find every critical point of $f$ that lies inside $(a, b)$.
Evaluate $f$ at each of those critical points.
Evaluate $f$ at both endpoints, $x=a$ and $x=b$.
The largest of these values is the absolute maximum; the smallest is the absolute minimum.
The endpoints are not optional. Because a closed interval includes its ends, the highest or lowest value can occur there even when no turning point does.
Example: absolute extrema of $f(x)=x^3-3x$ on $[0, 2]$.
The critical points are $x=-1$ and $x=1$, but only $x=1$ lies inside $(0, 2)$, so discard $x=-1$. Now evaluate at the interior critical point and both endpoints:
$$f(0) = 0, \qquad f(1) = 1 - 3 = -2, \qquad f(2) = 8 - 6 = 2$$
Compare the three values $0$, $-2$, and $2$.
Final answer: the absolute maximum is $2$ at the endpoint $x=2$, and the absolute minimum is $-2$ at $x=1$.
Notice that the absolute maximum landed on an endpoint, not on a turning point. A student who only tested critical points would report the wrong maximum. That is exactly why step 3 exists.
How Do You Solve An Optimization Problem With Maxima And Minima?
Real problems rarely hand you a function. Optimization means building the function first, then finding its maxima and minima. The pattern is always the same: name the variables, write the quantity to optimize, use a constraint to reduce it to one variable, differentiate, and verify.
Example: largest rectangular garden from 40 metres of fencing.
Let the rectangle have width $x$ and height $y$. The fence sets the perimeter:
$$2x + 2y = 40 \quad \Rightarrow \quad y = 20 - x$$
The area to maximize, written in one variable, is:
$$A(x) = x,y = x(20 - x) = 20x - x^2, \qquad 0 < x < 20$$
Differentiate and set to zero:
$$A'(x) = 20 - 2x = 0 \quad \Rightarrow \quad x = 10$$
Confirm it is a maximum with the second derivative:
$$A''(x) = -2 < 0 \quad \Rightarrow \quad \text{concave down, a maximum}$$
Then $y = 20 - 10 = 10$, and the area is $A(10) = 10 \times 10 = 100$ square metres. The best rectangle is a $10 \times 10$ square. The graph of $A(x)$ is a downward parabola, and its vertex is the maximum we just found.
Final answer: a $10\text{ m} \times 10\text{ m}$ square gives the largest area, $100$ square metres.
Which Test Should You Use For Maxima And Minima?
Three tools cover almost every problem. This table decides between them.
Table: Choosing the right method for a maxima and minima problem.
You want to find | Best method | What to check |
|---|---|---|
Local max/min, $f''$ easy to compute | Second derivative test | Sign of $f''(c)$ at each $f'(c)=0$ |
Local max/min, $f''(c)=0$ or messy | First derivative test | Sign change of $f'$ across $c$ |
Absolute max/min on $[a,b]$ | Closed-interval method | Critical points inside $(a,b)$ plus both endpoints |
Absolute max/min, real-world model | Optimization | Build the function, then apply a test above |
If the second derivative test returns $f''(c)=0$, do not guess. Switch to the first derivative test, which never fails to classify a genuine sign change.
Why Do The Derivative Tests Work?
The tests are not arbitrary recipes. They follow from one clean idea about smooth curves.
Fermat's insight. At a smooth interior peak or valley, the function stops rising and has not yet started falling, so its instantaneous rate of change is zero. That is Fermat's theorem: if $f$ has a local extremum at $c$ and is differentiable there, then $f'(c)=0$. It is why critical points are the only interior candidates worth checking.
The first test reads direction. The sign of $f'$ is the direction of travel: positive means uphill, negative means downhill. A peak is uphill-then-downhill, so $f'$ must switch $+$ to $-$; a valley is the reverse.
The second test reads bend. The sign of $f''$ is the concavity. A dome curving downward ($f''<0$) can only hold a maximum at its flat top; a cup curving upward ($f''>0$) can only hold a minimum at its base.
The one caution built into the second test is honest: when $f''(c)=0$, the curve is momentarily straight in its bending, and concavity alone cannot tell a peak from a valley from a mere pause. That is not a flaw, it is the test reporting the limit of its own information.
Who Discovered Maxima And Minima?
The method for finding maxima and minima is older than the derivative itself. A French lawyer working by candlelight found it decades before Newton and Leibniz built calculus.
Two names anchor the history:
Pierre de Fermat (1607–1665, France) gave the first general method for maxima and minima using adequality, the direct ancestor of setting $f'(x)=0$.
Sir Isaac Newton (1643–1727, England) and Gottfried Wilhelm Leibniz (1646–1716, Germany) later built the derivative into a full system, turning Fermat's trick into the routine first and second derivative tests taught today.
Where Are Maxima And Minima Used In The Real World?
The same peak-and-valley search runs under a wide range of real decisions.
Physics and motion: the maximum height of a projectile, and the moment a velocity is greatest or least, are extrema of position and velocity functions.
Economics: firms maximize profit and minimize cost by finding where marginal revenue equals marginal cost, which is a derivative set to zero.
Engineering: designers minimize material, weight, or stress while meeting a fixed constraint, exactly the optimization loop shown above.
Machine learning: training a model means minimizing a loss function, and gradient descent is a numerical hunt for that minimum.
Biology and medicine: peak drug concentration after a dose, and the fastest growth rate of a population, are located as maxima of the relevant curves.
One idea, "find where the rate of change is zero," quietly powers rocket trajectories, pricing models, structural design, and the training of large neural networks.
What Are The Most Common Mistakes With Maxima And Minima?
These four errors account for most lost marks, and each was confirmed against real student questions on Reddit's r/calculus and r/learnmath, Quora threads on the second derivative test, and AP Calculus review guides.
Forgetting the endpoints on a closed interval.
Where it slips in:
A student finds the absolute extrema on $[a,b]$ by testing only the critical points and never evaluates $f(a)$ and $f(b)$.
Don't do this:
Do not report the largest critical value as the absolute maximum. On a closed interval the extreme value can sit at an endpoint.
The correct way:
Evaluate $f$ at every interior critical point and at both endpoints, then compare the whole list. For $f(x)=x^3-3x$ on $[0,2]$, the maximum $2$ is at the endpoint $x=2$, not at any turning point.
Missing the critical points where $f'$ is undefined.
Where it slips in:
A student solves $f'(x)=0$ and stops, ignoring corners or vertical tangents where the derivative fails to exist.
Don't do this:
Do not treat "$f'(x)=0$" as the complete definition of a critical point.
The correct way:
Include every point where $f'$ is zero or undefined. The minimum of $f(x)=|x|$ is at $x=0$, where $f'(0)$ does not exist, and $f(x)=x^{2/3}$ has its minimum at the cusp $x=0$ for the same reason.
Misreading the second derivative test when $f''(c)=0$.
Where it slips in:
A student sees $f''(c)=0$ and concludes "inflection point" or "no extremum," treating the inconclusive result as an answer.
Don't do this:
Do not decide anything from $f''(c)=0$ alone. It does not confirm a maximum, a minimum, or an inflection.
The correct way:
When $f''(c)=0$, switch to the first derivative test. For $f(x)=x^4$, $f''(0)=0$ yet $x=0$ is a minimum; for $f(x)=x^3$, $f''(0)=0$ yet $x=0$ is not an extremum. Only the sign change of $f'$ settles it.
Thinking $f'(c)=0$ by itself means a maximum or minimum.
Where it slips in:
A student finds $f'(c)=0$ and immediately labels $c$ an extremum without checking whether the function actually turns there.
Don't do this:
Do not skip the classification step. A zero derivative marks a candidate, not a guaranteed peak or valley.
The correct way:
Confirm with a test. If $f'$ does not change sign across $c$ (as at $x=0$ for $y=x^3$), the point is a stationary inflection, not a maximum or minimum.
Practice Problems On Maxima And Minima
Work each one, then check the answer.
Find the critical points of $f(x)=x^2-4x+1$.
(Answer: $f'(x)=2x-4=0$, so $x=2$.)Classify $x=2$ for the function above.
(Answer: $f''(x)=2>0$, so $x=2$ is a local minimum; $f(2)=-3$.)Find the local extrema of $f(x)=x^3-12x$.
(Answer: $f'(x)=3x^2-12=0$ gives $x=\pm 2$; local max $f(-2)=16$, local min $f(2)=-16$.)Find the absolute extrema of $f(x)=x^2$ on $[-1, 3]$.
(Answer: critical point $x=0$ with $f(0)=0$; endpoints $f(-1)=1$, $f(3)=9$; absolute min $0$ at $x=0$, absolute max $9$ at $x=3$.)A rectangle has perimeter $24$. Maximize its area.
(Answer: $A(x)=x(12-x)$, $A'(x)=12-2x=0$, $x=6$; a $6\times 6$ square with area $36$.)For $f(x)=x^3$, show $x=0$ is not a local extremum.
(Answer: $f'(x)=3x^2\ge 0$ everywhere, so $f'$ never changes sign; $x=0$ is a stationary inflection, not an extremum.)
Where Should You Go Next After Maxima And Minima?
Maxima and minima open directly into the wider study of derivatives and their applications.
The derivative. Strengthen the tool every test depends on, the rate of change that goes to zero at a peak or valley.
Increasing and decreasing intervals. The sign-of-$f'$ reasoning behind the first derivative test, in its own right.
Tangents and the slope of a curve. The flat tangent at an extremum is where the geometry and the algebra meet.
If your child is learning to apply derivatives, a live Bhanzu trainer teaches maxima and minima from the geometry up, starting with the flat tangent and building to full optimization, in the Bhanzu math tutoring program.
Was this article helpful?
Your feedback helps us write better content