How do I write the equation of a tangent at a given point?

Two numbers are needed: the slope f′(x₀) and the value f(x₀). Put them into y = f′(x₀)(x − x₀) + f(x₀) and tidy up to get the form y = ax + b. The commonest mistake is computing f′(x₀) alone and forgetting that the line must also pass through the point of tangency.

How do I tell whether a function is increasing or decreasing?

From the sign of the derivative on an interval. Where f′(x) > 0 the tangents lean upwards and the function increases; where f′(x) 0 rather than checking individual values.

What is a critical point?

An argument at which the derivative equals zero (or fails to exist) — a place with a horizontal tangent. It is only a CANDIDATE for an extremum: the condition f′(x₀) = 0 is necessary but not sufficient, so every candidate still has to be tested.

Does f′(x₀) = 0 always mean there is an extremum?

No. The classic counterexample is f(x) = x³ at zero: the derivative 3x² vanishes there, but it is positive on both sides, so the function keeps increasing and there is no extremum — only a momentary flattening. An extremum needs the derivative to CHANGE sign.

How do I tell a maximum from a minimum?

By the direction in which the sign changes. If f′ goes from plus to minus, the function stops rising and starts falling — a maximum. From minus to plus — a minimum. It is convenient to record this in a sign table, with an arrow up or down under each interval.

How is a local extremum different from a largest value?

A local extremum is the highest (or lowest) point within some neighbourhood — the function may take even bigger values elsewhere. A largest value refers to the whole set under consideration. For f(x) = x³ − 3x the local maximum is 2, and yet the function reaches 18 at x = 3.

Is there a quicker way to tell a maximum from a minimum than a sign table?

Yes — the second derivative, the derivative of the derivative. If f′(x₀) = 0 and f″(x₀) 0 it is a minimum. When the second derivative vanishes too, the test decides nothing and you go back to the sign of the first derivative.

What does a typical optimisation problem look like?

Three steps: write the quantity you are optimising as a function of one variable, set its derivative to zero, then use the sign of the derivative to confirm that the point really is a maximum or a minimum. The first step is usually the hard one — the computation itself is mechanical.

Advanced level

What the derivative is for

The sign of the derivative says whether a function rises or falls, and its zeros point at the candidates for extrema. Learn to write the equation of a tangent, read a sign table and find a largest value.

Before you start

This topic builds on earlier ideas. Before you start, it's worth working through the lessons below — they'll make everything click:

The derivativeA derivative measures how fast a function changes at a point — it is the slope of the tangent to the graph. Meet the definition through the difference quotient and the six rules that compute it without taking a limit every time.
Linear functionA linear function y = ax + b draws a straight line. Meet the meaning of the slope and the intercept, the zero, the formula of a line through two points, the general form, and the conditions for parallel and perpendicular lines.

Where this is used

Real situations where you count exactly the way this lesson teaches:

Designing a drinks can
A can has to hold 330 ml on as little aluminium as possible. At radius r the height is h = 330/(πr²), so the surface area is S(r) = 2πr² + 660/r. The derivative S′(r) = 4πr − 660/r² vanishes at r³ = 660/(4π), that is r = 3.74 cm and h = 7.5 cm — the optimal can is squat, exactly as tall as it is wide. A real one has r = 3.3 cm and spends 268 cm² of metal instead of 264, because the lid must be thicker; across a billion cans that 1.5% is a budget line of its own.
Speed and fuel consumption
Fuel use does not simply fall with speed — it has a minimum. For the model f(v) = 0.0009v² − 0.126v + 9.3 litres per 100 km, the derivative f′(v) = 0.0018v − 0.126 vanishes at 70 km/h, where consumption is 4.89 l/100 km. At 120 km/h it is already 7.14 l, that is 46% more. The whole case for driving slower on a long trip fits into a number rather than a hunch, and it is the derivative that pins down where that point lies.
Order size in a warehouse
A shop sells 1,200 units a year; placing one order costs 90 in handling, and keeping one unit on the shelf costs 6 a year. The yearly cost is K(q) = 108,000/q + 3q, and K′(q) = −108,000/q² + 3 vanishes at q = 190 units. Ordering 100 every month costs 1,380 instead of 1,138 — 242 a year for a habit nobody ever costed. The minimum sits neither at the rarest nor at the most frequent deliveries, and without a derivative you would be groping for it.

All formulas

Equation of the tangent
$y = f'(x_0)(x - x_0) + f(x_0)$
a line of slope f′(x₀) through the point of tangency
Increasing function
$f'(x) > 0 \ \text{on an interval} \implies f \ \text{increasing}$
a positive derivative is a tangent leaning upwards
Decreasing function
$f'(x) < 0 \ \text{on an interval} \implies f \ \text{decreasing}$
a negative derivative is a tangent leaning downwards
Necessary condition for an extremum
$f'(x_0) = 0$
a critical point — a candidate, not yet an extremum
Sufficient condition (maximum)
$f' : + \to - \ \text{at} \ x_0 \implies \text{maximum at} \ x_0$
the derivative must CHANGE sign from plus to minus
Sufficient condition (minimum)
$f' : - \to + \ \text{at} \ x_0 \implies \text{minimum at} \ x_0$
minus to plus — a trough instead of a crest

The derivative is not an end in itself. Its value tells you how steeply a graph runs, and its sign tells you which way. Those two pieces of information are enough to read off the shape of a function without drawing it.

The equation of a tangent

A tangent at $x_0$ is a straight line, so a slope and one point describe it fully. The slope is the derivative and the point is $(x_0,\ f(x_0))$ :

y = f'(x_0)(x - x_0) + f(x_0)

A tangent touches the parabola at exactly one point and shares its slope there. Here f′(3) = 2, so the tangent climbs by 2 for every unit to the right.

Write the equation of the tangent to f(x) = x² − 4x + 5 at the point with x₀ = 3.

The sign of the derivative and monotonicity

Since the derivative is the slope of the tangent, its sign says outright which way the graph runs:

f'(x) > 0 \ \text{on an interval} \implies f \ \text{increasing there}

f'(x) < 0 \ \text{on an interval} \implies f \ \text{decreasing there}

This turns a question about monotonicity — which we used to read off a drawing — into an inequality. Instead of inspecting the graph, we solve $f'(x) > 0$ .

Mind the phrase "on an interval". Monotonicity is a property of an interval, never of a point, so the answer is always an interval and never a number.

Critical points

Between an interval where a function rises and one where it falls there has to be a place where it stops doing the first and starts doing the second. The derivative passes through zero there:

f'(x_0) = 0

Such an argument is called a critical point, and the condition is necessary for an extremum. Necessary, but not sufficient: it only produces a shortlist of candidates.

A counterexample worth remembering: for $f(x) = x^3$ the derivative $3x^2$ vanishes at zero, yet it is positive on both sides. The function keeps increasing and merely flattens for an instant — there is no extremum there.

The sufficient condition

An extremum requires the derivative to change sign:

f' : + \to - \ \text{at} \ x_0 \implies \text{local maximum at} \ x_0

f' : - \to + \ \text{at} \ x_0 \implies \text{local minimum at} \ x_0

At both critical points the tangent is horizontal. Left of x = −1 the function rises, between −1 and 1 it falls, and beyond it rises again — hence the crest and the trough.

Study the monotonicity and the extrema of f(x) = x³ − 3x.

Optimisation

The most practical use of a derivative: finding a largest or a smallest value. The pattern never changes — write the quantity as a function of one variable, set the derivative to zero, then check the sign to confirm which kind of extremum you found.

A ball thrown straight up has height h(t) = 20t − 5t² (metres, seconds). When does it reach its highest point, and how high is it?

Exercises

The three questions match the three steps of this lesson. For the tangent, type the right-hand side of $y = \ldots$ , that is an expression such as 2x - 4. For the zeros of the derivative, give both critical points in the separate $x_1$ and $x_2$ fields. For the extremum, the answer is its value, the height of the crest or the depth of the trough — not the argument at which it is reached.

Practice

Work through a set of exercises — they get harder as you go. At the end you'll see your score and the mistakes worth reviewing.

Exercise 1 of 8Score: 0

Tangent to the graph: f(x) = 2x² + x + 6, x₀ = 2 → y

Common mistakes

A tangent without its point of tangency — $f'(x_0)$ alone gives the slope; the line must still pass through $(x_0,\ f(x_0))$ .
An extremum from f′(x₀) = 0 alone — that condition is necessary; with no change of sign there is no extremum (witness $x^3$ at zero).
Giving the argument instead of the value of the extremum — an extremum is $f(x_0)$ , not $x_0$ .
Monotonicity at a point — a function increases on an interval, so the answer is an interval.
Confusing a local extremum with a largest value — "local" concerns a neighbourhood only.
Reading the sign on one side only — settling the question needs both intervals around the critical point.

Formula card

Topic: What the derivative is for

Equation of the tangent
$y = f'(x_0)(x - x_0) + f(x_0)$
a line of slope f′(x₀) through the point of tangency
Increasing function
$f'(x) > 0 \ \text{on an interval} \implies f \ \text{increasing}$
a positive derivative is a tangent leaning upwards
Decreasing function
$f'(x) < 0 \ \text{on an interval} \implies f \ \text{decreasing}$
a negative derivative is a tangent leaning downwards
Necessary condition for an extremum
$f'(x_0) = 0$
a critical point — a candidate, not yet an extremum
Sufficient condition (maximum)
$f' : + \to - \ \text{at} \ x_0 \implies \text{maximum at} \ x_0$
the derivative must CHANGE sign from plus to minus
Sufficient condition (minimum)
$f' : - \to + \ \text{at} \ x_0 \implies \text{minimum at} \ x_0$
minus to plus — a trough instead of a crest

The tangent to f(x) = x² − 4x + 5 at P(3, 2). Its equation y = 2x − 4 comes from the slope f′(3) = 2 and the coordinates of the point of tangency.

The graph of f(x) = x³ − 3x with two horizontal tangents. At x = −1 and x = 1 the derivative vanishes: the first is a local maximum of value 2, the second a local minimum of value −2.

Frequently asked questions

A luminous parabola and a straight line crossing above a faint coordinate grid, the vertex flaring like a star, against a cosmic nebula in violet and magenta

Functions and analysis

Why a function's graph can never double back

The whole definition of a function comes down to one word: uniqueness. One argument, one value — and not a single one more. That seemingly minor condition decides why a circle is not the graph of a function, why a line and a parabola describe so much of the world, and why four datasets with identical summary statistics can look nothing alike.

12 min