Math Core

You now have three tools for reading a function: $f$ tells you heights, $f'$ tells you where the graph rises and falls, and $f''$ tells you how it bends. In this lesson you'll put them together to sketch a graph from its formula, and to move back and forth between the graphs of $f$ , $f'$ and $f''$ the way AP questions ask you to.

How f, f′ and f″ fit together

Every feature of the graph of $f$ shows up as a feature of its derivatives.

feature of $f$	what $f'$ does	what $f''$ does
increasing	positive (above the axis)
decreasing	negative (below the axis)
relative max	changes from $+$ to $-$	often negative
relative min	changes from $-$ to $+$	often positive
concave up	increasing	positive
concave down	decreasing	negative
inflection point	has a relative max or min	changes sign

Shape from signs

The signs of $f'$ and $f''$ together determine the local shape of the graph:

$f' > 0$ , $f'' > 0$ : rising and bending up (increasing at an increasing rate).
$f' > 0$ , $f'' < 0$ : rising and bending down (increasing at a decreasing rate).
$f' < 0$ , $f'' > 0$ : falling and bending up (decreasing at a decreasing rate).
$f' < 0$ , $f'' < 0$ : falling and bending down (decreasing at an increasing rate).

A curve-sketching checklist

Domain, and any intercepts that are easy to find.
Asymptotes: vertical asymptotes where the function blows up, horizontal asymptotes from $\lim_{x \to \pm\infty} f(x)$ .
First derivative: critical points, increasing/decreasing intervals, relative extrema.
Second derivative: concavity intervals and inflection points.
Plot the key points and connect them with the right shape on each interval.

Worked example: A full analysis of a polynomial

Sketch $f(x) = 3x^4 - 4x^3$ .

Solution. The domain is all real numbers. The $x$ -intercepts come from $x^3(3x - 4) = 0$ : $x = 0$ and $x = \dfrac{4}{3}$ . There are no asymptotes, and $f(x) \to \infty$ as $x \to \pm\infty$ .

First derivative. $f'(x) = 12x^3 - 12x^2 = 12x^2(x - 1)$ . The critical points are $x = 0$ and $x = 1$ . Because $x^2 \ge 0$ , $f'$ has the sign of $x - 1$ : negative for $x < 1$ (except $0$ at $x = 0$ ), positive for $x > 1$ . So $f$ decreases for $x < 1$ and increases for $x > 1$ . There is a relative minimum at $(1, f(1)) = (1, -1)$ , and no extremum at $x = 0$ because $f'$ doesn't change sign there.

Second derivative. $f''(x) = 36x^2 - 24x = 12x(3x - 2)$ , which is zero at $x = 0$ and $x = \dfrac{2}{3}$ .

interval	$x < 0$	$0 < x < \frac{2}{3}$	$x > \frac{2}{3}$
sign of $f''$	$+$	$-$	$+$
concavity	up	down	up

Both are inflection points: $(0, 0)$ and $\left(\dfrac{2}{3}, -\dfrac{16}{27}\right) \approx (0.667, -0.593)$ .

f(x) = 3x⁴ − 4x³ (solid) with f′(x) = 12x²(x − 1) (dashed). The graph flattens at the origin without turning around, then bottoms out at (1, −1).Open in grapher →

At $x = 0$ the graph has a horizontal tangent and an inflection point: it levels off for an instant while switching from concave up to concave down, then keeps falling.

Worked example: A rational function with an asymptote

Sketch $f(x) = \dfrac{x^2}{x^2 + 3}$ .

Solution. The denominator is never $0$ , so the domain is all real numbers and there's no vertical asymptote. As $x \to \pm\infty$ , $f(x) \to 1$ , so $y = 1$ is a horizontal asymptote. $f$ is even (symmetric about the $y$ -axis), and $f(0) = 0$ .

First derivative. By the quotient rule,

f'(x) = \frac{2x(x^2 + 3) - x^2(2x)}{(x^2 + 3)^2} = \frac{6x}{(x^2 + 3)^2}.

$f' < 0$ for $x < 0$ and $f' > 0$ for $x > 0$ , so $f$ has a relative (in fact absolute) minimum at $(0, 0)$ .

Second derivative. Differentiating again and simplifying,

f''(x) = \frac{18(1 - x^2)}{(x^2 + 3)^3}.

$f'' > 0$ for $-1 < x < 1$ and $f'' < 0$ for $\lvert x \rvert > 1$ . So $f$ is concave up on $(-1, 1)$ , concave down outside, with inflection points at $\left(\pm 1, \dfrac{1}{4}\right)$ .

f(x) = x²/(x² + 3). It rises toward the horizontal asymptote y = 1, switching from concave up to concave down at x = ±1.Open in grapher →

Reading f from the graph of f′

When you're handed only the graph of $f'$ , read it in two passes:

Where is $f'$ above or below the axis? That gives increasing/decreasing and relative extrema of $f$ .
Where is $f'$ rising or falling? That gives concavity of $f$ , and the peaks and valleys of $f'$ are inflection points of $f$ .

Worked example: Analyzing f from f′

The graph of $f'$ is shown. Find the $x$ -coordinates of the relative extrema and inflection points of $f$ , and the interval where $f$ is concave up.

The graph of f′(x) = x² − 2x − 3 = (x + 1)(x − 3).Open in grapher →

Solution.

$f'$ changes from positive to negative at $x = -1$ , so $f$ has a relative maximum at $x = -1$ .
$f'$ changes from negative to positive at $x = 3$ , so $f$ has a relative minimum at $x = 3$ .
$f'$ decreases for $x < 1$ and increases for $x > 1$ , so $f$ is concave down for $x < 1$ , concave up for $x > 1$ , and has an inflection point at $x = 1$ .

Common mistake

When a problem shows the graph of $f'$ , its $x$ -intercepts are not the zeros of $f$ , and its peaks are not the peaks of $f$ . Say to yourself, "this is the slope," before answering. Many AP multiple-choice distractors are designed to catch students who read the graph of $f'$ as if it were $f$ .

Tip

To sketch $f'$ from the graph of $f$ , reverse the process: mark the $x$ -values where $f$ has horizontal tangents (those are zeros of $f'$ ), then decide whether $f'$ is positive or negative between them by asking whether $f$ is rising or falling.

Practice

The graph below shows $f'$ , the derivative of a function $f$ . Use it for the first three problems.

The graph of f′(x) = (x + 2)(x − 2)(x − 4)/4.Open in grapher →

Practice 1

Using the graph of $f'$ above, find all $x$ -values in $(-3.5, 5.5)$ where $f$ has a relative minimum.

Separate answers with commas, e.g. 2, -5

Practice 2

Using the graph of $f'$ above, on which interval is $f$ both increasing and concave down?

Practice 3

Using the graph of $f'$ above, how many inflection points does the graph of $f$ have on $(-3.5, 5.5)$ ?

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 4

Find the relative maximum value of $f(x) = \dfrac{x}{x^2 + 1}$ .

Enter a number. Fractions like 3/4 and sqrt(2) are OK.

Practice 5

Find the $x$ -coordinates of all inflection points of $f(x) = \dfrac{x}{x^2 + 1}$ .

Separate answers with commas, e.g. 2, -5

Practice 6

Find the $x$ -coordinates of all inflection points of $f(x) = x^4 - 4x^3$ .

Separate answers with commas, e.g. 2, -5

Practice 7

A differentiable function $g$ satisfies $g'(x) < 0$ and $g''(x) > 0$ for all $x$ in $(0, 5)$ . Which describes the graph of $g$ on that interval?