Curve Sketching: Connecting f, f′ and f″

Q: What does it mean if f′ > 0 and f″ < 0?

The graph is rising but bending downward, so it rises more and more slowly.

Q: Is a maximum of f′ a maximum of f?

No. A maximum of f′ marks the steepest point of f, which is an inflection point, not a peak.

Quick answer

A graph's shape is written in its derivatives: f′ tells where it rises and falls and where it peaks, and f″ tells how it bends and where the bend changes. A full sketch gathers the domain, intercepts, symmetry and asymptotes, then the sign charts of f′ and f″, then the key points. The same connections run backward: from a graph of f′, f increases where f′ is above the axis, has extrema where f′ crosses it, and has inflection points where f′ turns around.

What you'll learn

Assemble domain, intercepts, symmetry, asymptotes and derivative information into a sketch
Describe a graph's shape from the signs of f′ and f″
Read the behavior of f from a graph of f′

The checklist

To sketch $y = f(x)$ from its formula:

Domain, intercepts and symmetry (even: $f(-x) = f(x)$ ; odd: $f(-x) = -f(x)$ ).
Asymptotes: vertical ones where $f$ blows up, horizontal ones from the limits at $\pm\infty$ .
First derivative: critical points, a sign chart, where $f$ increases and decreases, local extrema.
Second derivative: a sign chart, concavity, inflection points.
Plot the key points and draw a curve that matches every sign.

Four shapes from two signs

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

	$f'' > 0$ (concave up)	$f'' < 0$ (concave down)
$f' > 0$	rising, faster and faster	rising, slower and slower
$f' < 0$	falling, slower and slower	falling, faster and faster

Why the derivatives determine the shape

$f'$ is the slope at every point, so it fixes the direction of the curve. $f''$ is how the slope changes, so it fixes the bend. Between the points where either one changes sign, the curve cannot turn around or change its bend — those are exactly the events the sign charts rule out.

So the graph is pinned down in pieces, and the key points — extrema, inflection points, intercepts — are where the pieces join. Once you know every place the slope or the bend changes, only one shape fits between them.

Example 1 — A complete sketch

Sketch $f(x) = \tfrac{1}{4}x^4 - 2x^2 + 3$ .

The domain is all reals, $f(0) = 3$ , and $f$ is even, so the graph is symmetric about the $y$ -axis. As a polynomial it has no asymptotes; it grows without bound at both ends.

$f'(x) = x^3 - 4x = x(x - 2)(x + 2)$ . Signs on the four intervals: $-$ , $+$ , $-$ , $+$ . So $f$ has local minima at $x = \pm 2$ , where $f = -1$ , and a local maximum at $x = 0$ , where $f = 3$ .

$f''(x) = 3x^2 - 4$ is zero at $x = \pm\tfrac{2}{\sqrt{3}} \approx \pm 1.15$ and changes sign there: inflection points at about $(\pm 1.15, 0.78)$ .

f(x) = ¼x⁴ − 2x² + 3

A sketch from the derivatives

Reading f from a graph of f′

Often only the graph of $f'$ is given. Read it this way:

On the graph of $f'$	For $f$
above the $x$ -axis	$f$ increasing
below the $x$ -axis	$f$ decreasing
crosses from $+$ to $-$	local maximum of $f$
crosses from $-$ to $+$	local minimum of $f$
$f'$ increasing	$f$ concave up
$f'$ has a local max or min	inflection point of $f$

f
f′

A function and its derivative

Worked examples

Example 2 — A rational function

Sketch $g(x) = \tfrac{x}{x^2 + 1}$ .

The domain is all reals, the only intercept is $(0, 0)$ , and $g$ is odd. Since the denominator has the higher degree, $y = 0$ is a horizontal asymptote.

$g'(x) = \tfrac{1 - x^2}{(x^2 + 1)^2}$ : negative, positive, negative around $-1$ and $1$ . So there is a local minimum $\left(-1, -\tfrac{1}{2}\right)$ and a local maximum $\left(1, \tfrac{1}{2}\right)$ .

$g''(x) = \tfrac{2x(x^2 - 3)}{(x^2 + 1)^3}$ changes sign at $0$ and $\pm\sqrt{3}$ : inflection points at $(0, 0)$ and $\left(\pm\sqrt{3}, \pm\tfrac{\sqrt{3}}{4}\right)$ .

Common mistakes

Practice problems

Find the local extrema and inflection points of $f(x) = x^3 - 3x$ .

Answer

Local max $(-1, 2)$ ; local min $(1, -2)$ ; inflection point $(0, 0)$

Full solution

$f'(x) = 3x^2 - 3$ , zero at $\pm 1$ , with signs $+$ , $-$ , $+$ . $f''(x) = 6x$ changes sign at $0$ .
Describe the shape of a graph with $f' > 0$ and $f'' < 0$ on an interval.

Answer

Rising, but more and more slowly

Full solution

$f' > 0$ means rising; $f'' < 0$ means the slope is decreasing, so the rise flattens out.
$f'$ is positive on $(-\infty, 3)$ and negative on $(3, \infty)$ . What happens at $x = 3$ ?

Answer

$f$ has a local maximum.

Full solution

$f$ increases up to $3$ and decreases after it.
The graph of $f'$ has a local minimum at $x = 1$ . What does $f$ have there?

Answer

An inflection point

Full solution

$f'$ switches from decreasing to increasing at $1$ , so $f''$ changes from negative to positive: the concavity of $f$ changes.
Find the asymptotes of $h(x) = \tfrac{x^2 - 4}{x^2 - 1}$ .

Answer

Vertical: $x = \pm 1$ . Horizontal: $y = 1$ .

Full solution

The denominator is $0$ at $\pm 1$ and the numerator is not, so those are vertical asymptotes. Equal degrees give the horizontal asymptote $y = \tfrac{1}{1} = 1$ .
Is $f(x) = x^4 - 2x^2$ even, odd or neither? What does that mean for its graph?

Answer

Even; the graph is symmetric about the $y$ -axis.

Full solution

$f(-x) = x^4 - 2x^2 = f(x)$ , so anything found for $x > 0$ is mirrored for $x < 0$ .
Where is $f(x) = x e^{x}$ increasing, and where is it concave up?

Answer

Increasing on $(-1, \infty)$ ; concave up on $(-2, \infty)$

Full solution

$f'(x) = e^x(x + 1)$ is positive for $x > -1$ . $f''(x) = e^x(x + 2)$ is positive for $x > -2$ .
A graph of $f'$ lies above the $x$ -axis everywhere and is increasing. Describe $f$ .

Answer

Always increasing and always concave up

Full solution

$f' > 0$ gives increasing, and $f'$ increasing means $f'' > 0$ : concave up.
Find the inflection points of $f(x) = x^4 - 6x^2$ .

Answer

$(-1, -5)$ and $(1, -5)$

Full solution

$f''(x) = 12x^2 - 12$ , which changes sign at $\pm 1$ . $f(\pm 1) = 1 - 6 = -5$ .
Given the graph of $f'$ , a student marks the highest point of that graph as the maximum of $f$ . What went wrong?

Hint

What does a high value of $f'$ say about $f$ ?

Answer

The highest point of $f'$ is where $f$ is steepest, an inflection point. The maxima of $f$ are where $f'$ crosses from positive to negative.

Full solution

$f'$ is largest where $f$ climbs fastest. There $f'$ turns from increasing to decreasing, so the concavity of $f$ changes.

A maximum of $f$ needs $f'$ to change sign, which happens where the graph of $f'$ crosses the axis.

Frequently asked questions

What are the steps of curve sketching?

Domain, intercepts and symmetry; asymptotes; a sign chart of f′ for increasing, decreasing and extrema; a sign chart of f″ for concavity and inflection points; then plot the key points and connect them to match.

How do I find where f is increasing from a graph of f′?

Look for where the graph of f′ lies above the x-axis. f is increasing there, and decreasing where f′ lies below.

Where does f have an inflection point, judging from the graph of f′?

Where f′ has a local maximum or minimum, since that is where f′ switches between increasing and decreasing, so f″ changes sign.

What does it mean if f′ > 0 and f″ < 0?