Curve Sketching

Curve sketching means drawing an accurate graph of a function from its formula, using f, f′ and f″ together. From f you get the domain, intercepts, symmetry and asymptotes. From f′ you get where the graph rises and falls and its local extrema. From f″ you get concavity and inflection points.

Before this lesson: First Derivative Test, Concavity and Inflection Points and Limits at Infinity

Why use calculus to sketch a graph

Plotting a few points can miss important features: a hidden peak between two points, or the exact place where a curve switches from bending up to bending down. The derivatives tell you exactly where the turning points and bends are, so the sketch has the right shape everywhere.

Curve sketching checklist

Not every function needs every step. A polynomial has no asymptotes, for example. A summary table that lists each interval with the signs of f′ and f″ keeps all the information in one place.

Curve sketching examples

Example 1: A cubic

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

x^3 - 3x = x(x^2 - 3) = 0 \Rightarrow x = 0, \pm\sqrt{3}

f(-x) = -x^3 + 3x = -f(x)

f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1)

f(-1) = 2, \; f(1) = -2

f''(x) = 6x

f(0) = 0

Answer: the graph rises to (−1, 2), falls through the inflection point (0, 0) to (1, −2), then rises again, crossing the x-axis at about ±1.73. It is concave down on the left half and concave up on the right half.

Example 2: A quartic

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

x^3(x - 4) = 0 \Rightarrow x = 0, 4

f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3)

f(3) = 81 - 108 = -27

f''(x) = 12x^2 - 24x = 12x(x - 2)

f(0) = 0, \; f(2) = 16 - 32 = -16

Answer: the graph comes down from the upper left, flattens at the origin (an inflection point with a horizontal tangent), keeps falling, changes concavity at (2, −16), reaches its minimum at (3, −27), and rises through (4, 0).

Example 3: A rational function with a horizontal asymptote

Sketch f(x) = \frac{x}{x^2 + 1}.

\lim_{x \to \pm\infty}\frac{x}{x^2 + 1} = 0

f(-x) = -f(x)

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

f(-1) = -\frac{1}{2}, \; f(1) = \frac{1}{2}

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

f(\pm\sqrt{3}) = \pm\frac{\sqrt{3}}{4} \approx \pm 0.433

Answer: the graph hugs y = 0 on the far left, dips to (−1, −½), rises through the origin to (1, ½), then sinks back toward y = 0. It has three inflection points.

Example 4: A rational function with vertical asymptotes

Sketch f(x) = \frac{x^2}{x^2 - 4}.

x^2 - 4 = 0 \Rightarrow x = \pm 2

\lim_{x \to \pm\infty}\frac{x^2}{x^2 - 4} = 1

f'(x) = \frac{-8x}{(x^2 - 4)^2}

f(0) = 0

f''(x) = \frac{8(3x^2 + 4)}{(x^2 - 4)^3}

\text{up for } |x| \gt 2, \; \text{down for } |x| \lt 2

Answer: three pieces. Outside the asymptotes the graph is concave up and approaches y = 1 from above. Between them it is a cap with its top at (0, 0), falling to −∞ at both asymptotes.

Example 5: Word problem with a medicine

The concentration of a medicine in the blood t hours after a pill is taken is C(t) = 10te^{-t/2} mg/L, for t \ge 0. Describe its graph: when is the concentration highest, and when is it falling fastest?

C'(t) = 10e^{-t/2} - 5te^{-t/2} = 5e^{-t/2}(2 - t)

C(2) = 20e^{-1} \approx 7.36

C''(t) = \frac{5}{2}e^{-t/2}(t - 4)

C(4) = 40e^{-2} \approx 5.41

\lim_{t \to \infty} C(t) = 0

Answer: the concentration rises to its peak of about 7.36 mg/L at 2 hours. It falls fastest at the inflection point, 4 hours (about 5.41 mg/L), and then levels off toward 0.

Common mistakes

Related methods

Sketching from a graph of f′

You can also work backward: given a graph of f′, the x-intercepts of f′ show where f has horizontal tangents, and the slope of f′ gives the concavity of f. This reverses sketching the graph of a derivative.

Checking with technology

A graphing calculator is a good way to check a sketch, but the analysis tells you why the graph looks the way it does and where the exact turning points are, which a screen can hide.

Practice problems

1. Sketch f(x) = x^2 - 4x: give the intercepts, extrema and concavity.
2. Sketch f(x) = x^3 - 12x: give the extrema and the inflection point.
3. Sketch f(x) = x^3 - 3x^2: give the intercepts, extrema and inflection point.
4. Sketch f(x) = x^4 - 2x^2: give the extrema and inflection points.
5. Sketch f(x) = \frac{1}{x^2 + 1}: give the asymptote, extremum and inflection points.
6. Sketch f(x) = \frac{x + 1}{x - 1}: give the asymptotes, where it increases or decreases, and its concavity.
7. Sketch f(x) = xe^x: give the extremum, inflection point and asymptote.
8. Sketch f(x) = x - \ln x: give the domain, extremum and concavity.
9. Sketch f(x) = x + \sin x on [0, 2π]: where is it increasing, and where is its inflection point?
10. A population is P(t) = \frac{100t^2}{t^2 + 25} (in thousands) after t years. Find its long-run value and when it grows fastest.

x(x - 4) = 0 \Rightarrow x = 0, 4

f'(x) = 2x - 4 \Rightarrow x = 2, \; f(2) = -4

f''(x) = 2 \gt 0

f'(x) = 3x^2 - 12 \Rightarrow x = \pm 2

f(-2) = 16, \; f(2) = -16

f''(x) = 6x \Rightarrow (0, 0)

x^2(x - 3) = 0 \Rightarrow x = 0, 3

f'(x) = 3x^2 - 6x = 3x(x - 2)

f''(x) = 6x - 6 \Rightarrow x = 1, \; f(1) = -2

f'(x) = 4x(x - 1)(x + 1)

f(0) = 0, \; f(\pm 1) = -1

f''(x) = 12x^2 - 4 = 0 \Rightarrow x = \pm\frac{1}{\sqrt{3}}

f\left(\pm\frac{1}{\sqrt{3}}\right) = \frac{1}{9} - \frac{2}{3} = -\frac{5}{9}

\lim_{x \to \pm\infty}\frac{1}{x^2 + 1} = 0

f'(x) = \frac{-2x}{(x^2 + 1)^2} \Rightarrow (0, 1)

f''(x) = \frac{6x^2 - 2}{(x^2 + 1)^3} = 0 \Rightarrow x = \pm\frac{1}{\sqrt{3}}

x = 1, \; y = 1

f'(x) = \frac{-2}{(x - 1)^2} \lt 0

f''(x) = \frac{4}{(x - 1)^3}

f'(x) = (x + 1)e^x \Rightarrow x = -1

f''(x) = (x + 2)e^x \Rightarrow x = -2

\lim_{x \to -\infty} xe^x = 0

x \gt 0

f'(x) = 1 - \frac{1}{x} \Rightarrow x = 1, \; f(1) = 1

f''(x) = \frac{1}{x^2} \gt 0

f'(x) = 1 + \cos x \ge 0

f''(x) = -\sin x = 0 \Rightarrow x = \pi

f(\pi) = \pi

\lim_{t \to \infty}\frac{100t^2}{t^2 + 25} = 100

P'(t) = \frac{5000t}{(t^2 + 25)^2} \gt 0

P''(t) = \frac{5000(25 - 3t^2)}{(t^2 + 25)^3} = 0 \Rightarrow t = \frac{5}{\sqrt{3}} \approx 2.89

Curve sketching FAQ

What are the steps of curve sketching?

Find the domain, intercepts, symmetry and asymptotes; use f′ for increasing, decreasing and extrema; use f″ for concavity and inflection points; then plot the key points and connect them with the right shape.

What does the second derivative tell you about a graph?

Its sign gives the concavity: positive means the graph bends up like a cup, negative means it bends down like a cap.

Do I need a table of values?

Only for a few key points: intercepts, extrema and inflection points. The derivatives tell you the shape between them.

Can a graph cross its horizontal asymptote?

Yes. A horizontal asymptote describes the far ends of the graph only. In the middle, the graph can cross it.

Summary

• From f: domain, intercepts, symmetry, asymptotes.
• From f′: increasing and decreasing intervals, local max and min.
• From f″: concavity and inflection points.
• Plot the key points, then use the signs of f′ and f″ to draw the right shape between them.

Previous lesson: Second Derivative Test

Next lesson: Optimization

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