Sketching Graphs of Functions and Their Derivatives

AP Calculus · Analytical Applications of Differentiation

Sketching Graphs of Functions and Their Derivatives

Collect the facts first. The curve that fits them is essentially unique.

key idea$\text{Zeros of }f'\to\text{extrema of }f;\quad\text{zeros of }f''\to\text{inflections of }f$

Reading the graph

Sketching is assembly, not invention. Collect a short list of facts — intercepts, critical points, concavity, end behaviour — and the curve has essentially one shape that fits them all.

$f(x)=x^{3}-3x$

$f'(x)=3x^{2}-3$

$f$ and its derivative. Each zero of $f'$ sits directly below an extremum of $f$; where $f'$ is below the axis, $f$ falls. Reading vertically between the two panels is the whole skill.

combined sign chart for $f(x)=x^{3}-3x$

	$x\lt -1$	$-1\lt x\lt 0$	$0\lt x\lt 1$	$x\gt 1$
$f'$	+	−	−	+
$f''$	−	−	+	+
verdict	rising, cap	falling, cap	falling, cup	rising, cup

Feature of $f$	Comes from	Condition
$x$-intercepts	$f$	$f(x)=0$
$y$-intercept	$f$	evaluate $f(0)$
local extrema	$f'$	$f'=0$ with a sign change
intervals of increase	$f'$	$f'\gt 0$
inflection points	$f''$	$f''=0$ with a sign change
concavity	$f''$	sign of $f''$
end behaviour	$f$	limits as $x\to\pm\infty$
asymptotes	$f$	where $f$ is undefined or unbounded

Eight facts, three sources. Gather them before drawing anything — a sketch built fact-by-fact rarely needs correcting.

Worked examples

EX 1

Assemble a full sketch

$f(x)=x^{3}-3x$

Sketch $f(x)=x^{3}-3x$ using all available information.

Intercepts

$f(x)=x\left(x^{2}-3\right)=0$ at $x=0,\pm\sqrt3$. The $y$-intercept is $f(0)=0$.

Critical points

$f'=3x^{2}-3=0$ at $x=\pm1$. Sign: $+,-,+$, so a maximum at $(-1,2)$ and a minimum at $(1,-2)$.

Concavity

$f''=6x$, zero at $x=0$ with a sign change, so an inflection point at $(0,0)$; concave down for $x\lt 0$, up for $x\gt 0$.

End behaviour

As $x\to\infty$, $f\to\infty$; as $x\to-\infty$, $f\to-\infty$ — the usual odd-degree pattern.

Draw

Rise to $(-1,2)$, fall through the origin to $(1,-2)$, then rise. Every feature was determined before the pen moved.

EX 2

Sketch $f'$ from $f$

graph → graph

Given the graph of $f$ in the figure, sketch $f'$.

Mark where $f$ has horizontal tangents

At $x=-1$ and $x=1$. These become zeros of $f'$.

Read the direction between them

$f$ rises on $x\lt -1$, so $f'\gt 0$ there. It falls on $(-1,1)$, so $f'\lt 0$. It rises again after.

Find the extremum of $f'$

$f$ is steepest downward at $x=0$, so $f'$ has its minimum there.

Sketch

A parabola opening upward, crossing at $\pm1$, with vertex at $x=0$. That matches the right-hand panel.

The translation rule

Extremum of $f$ → zero of $f'$. Inflection of $f$ → extremum of $f'$. Degree drops by one.

EX 3

Sketch $f$ from $f'$

the harder direction

$f'$ is positive on $(-\infty,2)$, negative on $(2,5)$, positive on $(5,\infty)$, with $f(2)=7$. Describe $f$.

Read the direction

$f$ increases up to $x=2$, decreases on $(2,5)$, then increases again.

Classify the turning points

$+\to-$ at $x=2$ gives a local maximum; $-\to+$ at $x=5$ gives a local minimum.

Place what you know

The maximum is at $(2,7)$. The minimum's height is unknown — $f'$ alone cannot supply it.

Note the ambiguity

Any vertical shift of the sketch is consistent with the same $f'$. One value of $f$ is needed to pin it down, and $f(2)=7$ does that.

The asymmetry

Going from $f$ to $f'$ is determined. Going from $f'$ to $f$ leaves a constant free — the seed of antidifferentiation.

EX 4

Spot the inconsistency

checking a sketch

A student's sketch shows $f$ with a local maximum at $x=3$, yet their sign chart has $f'\gt 0$ on both sides of $3$. What is wrong?

Read the sign chart

$f'\gt 0$ on both sides means $f$ is increasing through $x=3$.

Read the sketch

A local maximum requires $f$ to rise then fall — $f'$ must change from $+$ to $-$.

Name the contradiction

The two cannot both hold. An increasing function has no interior local maximum.

Diagnose

Either the sign chart missed a sign change, or the sketch invented a turning point that the derivative does not support.

The habit

Cross-check the sketch against the sign chart before finishing. They must agree at every interval, and disagreement always means an error.

Watch out

Gather facts before drawing

A sketch drawn first and justified afterwards usually contains a feature no derivative supports — Example 4.

Zero of $f'$ is not automatically an extremum

The sign must change. A stationary point of inflection has $f'=0$ and no turning.

$f'$ cannot give you heights

It determines shape, not vertical position. One known value of $f$ is needed to fix the curve.

Read vertically between stacked graphs

A feature of $f$ at $x=a$ appears directly above or below the corresponding feature of $f'$ at the same $a$. The horizontal position is shared.

Course map. Topic 5.8 · LO FUN-4.A — justify conclusions about the graph of a function based on the behaviour of its derivatives (EK FUN-4.A.9, FUN-4.A.10).