Curve sketching

The first derivative tells you where a graph rises and falls, the second tells you how it bends, and limits at infinity tell you what it settles towards. Put together they let you draw a curve you have never seen, and read one you have.

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In this topic

Taught: Before Quiz 2
Questions: 29
Total marks: 71
Suggested time: 124 min
Difficulty: Introductory to Capstone
Show the 7 hardest

Key results

Increasing and decreasing: $\begin{aligned} f' > 0 \text{ on } (a, b) &\implies f \text{ increasing on } [a, b] \\ f' < 0 \text{ on } (a, b) &\implies f \text{ decreasing on } [a, b] \end{aligned}$
When $f$ is continuous on $[a, b]$ .
The first derivative test: $\begin{aligned} f' \colon + \text{ to } - \text{ at } c &\implies \text{local max at } c \\ f' \colon - \text{ to } + \text{ at } c &\implies \text{local min at } c \end{aligned}$
When the sign of $f'$ changes at $c$ as shown, read from left to right, and $f$ is continuous at $c$ , even if $f'(c)$ does not exist.
Concavity: $\begin{aligned} f'' > 0 \text{ on } I &\implies \text{concave up on } I \\ f'' < 0 \text{ on } I &\implies \text{concave down on } I \end{aligned}$
For an interval $I$ .
Inflection point: $\bigl(c, f(c)\bigr) \text{ is an inflection point} \iff \text{the concavity of } f \text{ changes at } c$
When $f$ is continuous at $c$ ; then $f''(c) = 0$ or $f''(c)$ does not exist, but $f''(c) = 0$ alone is not enough, as $x^{4}$ at 0 shows.
Asymptotes: $\begin{aligned} x = a&:\ \lim_{x \to a^{\pm}} f(x) = \pm\infty \\ y = L&:\ \lim_{x \to \pm\infty} f(x) = L \\ y = mx + b&:\ \lim_{x \to \pm\infty}\bigl(f(x) - (mx + b)\bigr) = 0 \end{aligned}$
Vertical, horizontal and slant in turn, each needing its limit from one side only ( $x \to a^{-}$ or $a^{+}$ ; $x \to \infty$ or $-\infty$ ), and for the vertical one either infinite sign.
Symmetry: $f(-x) = f(x) \ \text{(even)}, \qquad f(-x) = -f(x) \ \text{(odd)}, \qquad f(x + P) = f(x) \ \text{(periodic)}$
For every $x$ in the domain, with $P > 0$ ; the smallest such $P$ , when there is one, is the period.

Read the theoryUBC Math 100 §7 Sketching graphsCLP-1 §3.6 Sketching graphs

Worked example

Standard2 marks · 4 min

Find the local maximum value of $f(x) = x^{3} - 3x^{2} - 9x + 5$ .

Give the value of the function, not the x-coordinate.

Strategy

Locate the critical points, classify them with the first or second derivative test, then evaluate the original function at the one that is a maximum.

Working

$f'(x) = 3x^{2} - 6x - 9 = 3(x - 3)(x + 1)$
Differentiate and factor. Critical points at x = −1 and x = 3.
$f''(x) = 6x - 6$
The second derivative test is quicker than a sign chart here.
$f''(-1) = -12 < 0 \implies \text{local maximum at } x = -1$
Negative second derivative means concave down, so this critical point is a peak.
$f(-1) = -1 - 3 + 9 + 5 = 10$
Evaluate the original function, not the derivative, to get the value.

Answer

f(-1) = 10

Check

f(−1.1) = 9.939 and f(−0.9) = 9.941 both fall just short of f(−1) = 10.

Where this goes wrong

Answering −1, which is where the maximum occurs rather than what the maximum value is. Read carefully whether the question wants the location or the value.

On the paper

Show the domain, intercepts, asymptotes with the limits that give them, and sign charts for $f'$ and $f''$ . Label every extreme, inflection point and asymptote on the sketch.

Leans on

Differentiating a polynomial, from Differentiation rules

Your turn

Standard3 marks · 4 min

Find the $x$ -coordinate of the local maximum of $f(x) = xe^{-2x}$ .

Answer this question

Practice questions

5 of the 29, across every level. Open a solution to read the working, or answer the question yourself and it counts towards your progress.

Introductory2 marks · 3 min
Find the largest open interval on which $f(x) = \dfrac{e^{x}}{x + 2}$ is increasing.
Strategy
Differentiate, throw away the factors that are always positive, and read the sign of what remains.
Working
$f'(x) = \frac{e^{x}(x + 2) - e^{x}}{(x + 2)^{2}} = \frac{e^{x}(x + 1)}{(x + 2)^{2}}$
Quotient rule, then factor eˣ.
$f'(x) > 0 \iff x + 1 > 0 \iff x > -1$
eˣ > 0 and (x + 2)² > 0 wherever f is defined.
Answer
$(-1, \infty)$
The answer is x greater than negative 1.
Check
f(−1.5) ≈ 0.45, f(−1) ≈ 0.37 and f(0) = 0.5, so f is still falling just left of −1 and rising after it.
Where this goes wrong
Answering (−2, ∞) because that is where f is defined and positive. Being positive and being increasing are different; between −2 and −1 the function is positive but falling.
Leans on
The quotient rule, from Differentiation rules
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What is the largest open interval on which $f(x) = xe^{x}$ is increasing?
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Introductory1 mark · 2 min
Find the $x$ -coordinate of the inflection point of $f(x) = x^{3} - 6x^{2} + x$ .
Strategy
Concavity is governed by the second derivative. For a cubic, f″ is linear, so it changes sign exactly once, at its zero.
Working
$f'(x) = 3x^{2} - 12x + 1, \qquad f''(x) = 6x - 12$
Two derivatives.
$f''(x) = 0 \iff x = 2;\quad f'' < 0 \text{ for } x < 2,\ f'' > 0 \text{ for } x > 2$
Concave down then concave up: a genuine inflection point at x = 2.
Answer
$x = 2$
The answer is 2.
Check
The roots of x³ − 6x² + x add to 6, and the inflection point of a cubic sits at the average of its three roots, 6/3 = 2.
Where this goes wrong
Setting f′ = 0 instead of f″ = 0, which finds the local extrema rather than the inflection point.
Leans on
Higher-order derivatives, from Differentiation rules
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On which interval is $f(x) = x^{4} - 6x^{2}$ concave down?
Answer this question
Standard2 marks · 4 min
On which interval is $f(x) = x^{4} - 6x^{2}$ concave down?
Strategy
Differentiate twice, find where the second derivative is zero, and test the sign on each resulting interval. Concave down means the second derivative is negative.
Working
$f'(x) = 4x^{3} - 12x$
First derivative. Not what determines concavity, but needed to get to the second.
$f''(x) = 12x^{2} - 12 = 12(x - 1)(x + 1)$
Second derivative, factored so the sign changes are visible.
$f''(x) < 0 \iff (x - 1)(x + 1) < 0 \iff -1 < x < 1$
(x − 1)(x + 1) is negative exactly when its two factors have opposite signs, which is between −1 and 1.
$\text{Concave down on } (-1, 1)$
The points x = ±1 are inflection points, where the concavity changes.
Answer
$(-1, 1)$
The answer is the interval from −1 to 1.
Check
f(0) = 0 sits above the chord from x = −0.5 to 0.5, at −1.4375, but f(1.5) = −8.4375 sits below the chord from f(1) = −5 to f(2) = −8.
Where this goes wrong
Solving f′(x) = 0 instead of f″(x) = 0, which gives x = 0 and x = ±√3 and leads to the interval (−√3, √3). The first derivative describes rising and falling; the second describes bending.
Leans on
Sign charts, from Functions, graphs and inverses
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Find the $x$ -coordinate of the inflection point of $f(x) = x^{3} - 6x^{2} + x$ .
Answer this question
Challenging3 marks · 5 min
A function $f$ , defined for all real $x$ , has derivative $f'(x) = (3 - x)(x + 1)e^{-x}$ . You are given $f'$ , not $f$ . At which value of $x$ does $f$ have a local maximum?
Strategy
The first-derivative test needs only the sign of f′. The exponential factor is always positive, so the sign is decided by the two linear factors.
Working
$f'(x) = 0 \iff x = 3 \text{ or } x = -1$
Critical points; e^−x never vanishes.
$x < -1:\ (+)(-)(+) < 0;\qquad -1 < x < 3:\ (+)(+)(+) > 0;\qquad x > 3:\ (-)(+)(+) < 0$
Sign chart of (3 − x), (x + 1), e^−x.
$\text{local min at } x = -1,\quad \text{local max at } x = 3$
f′ goes − to + at −1 and + to − at 3.
Answer
$x = 3$
The answer is 3.
Check
f″(3) = −4e⁻³ is negative, so the second derivative test also gives a local maximum at x = 3.
Where this goes wrong
Differentiating f′ again and setting f″ = 0, as if looking for an inflection point, or trying to integrate to recover f. The question only needs the sign of the derivative provided.
Try a similar question
Find the $x$ -coordinate of the local maximum of $f(x) = xe^{-2x}$ .
Answer this question
Capstone6 marks · 12 min
Let $f(x) = \dfrac{x^{2}}{x - 1}$ . Find its domain and asymptotes (vertical and slant), the intervals on which it increases and decreases, its local extrema, and the intervals on which it is concave up and concave down. Then sketch the graph.
Strategy
Divide to reveal the linear part plus a simple remainder; every feature of the graph then follows from that remainder: where it blows up, where it flattens the slope, and which way it bends.
Working
$f(x) = x + 1 + \frac{1}{x - 1}, \qquad x \neq 1$
Division. As x → ±∞ the remainder → 0, so y = x + 1 is a slant asymptote; as x → 1⁻, f → −∞ and as x → 1⁺, f → +∞.
$f'(x) = 1 - \frac{1}{(x - 1)^{2}} = \frac{x(x - 2)}{(x - 1)^{2}}$
Zero at x = 0 and x = 2. Positive outside [0, 2], negative between (excluding 1).
$\text{local max } (0, 0),\qquad \text{local min } (2, 4)$
f′ changes + to − at 0 and − to + at 2. Note the “maximum” value 0 is below the “minimum” value 4: the asymptote sits between them.
$f''(x) = \frac{2}{(x - 1)^{3}}:\quad f'' < 0 \text{ for } x < 1,\quad f'' > 0 \text{ for } x > 1$
Concave down on the left branch, up on the right; no inflection point because concavity only changes across the asymptote.
Sketch: the left branch rises from the asymptote y = x + 1 far to the left, peaks at the origin, and dives to −∞ at x = 1; the right branch falls from +∞ at x = 1 to (2, 4) and then rises alongside y = x + 1.
Answer
$\text{VA } x = 1;\quad \text{SA } y = x + 1;\quad \text{increasing on } (-\infty, 0)\cup(2, \infty);\quad \text{decreasing on } (0, 1)\cup(1, 2);\quad \max (0,0),\ \min (2,4);\quad \text{concave down for } x<1,\ \text{up for } x>1$
The answer is vertical asymptote x = 1, slant asymptote y = x + 1, increasing left of 0 and right of 2, decreasing between (except at 1), local max at the origin, local min at (2, 4), concave down left of 1 and up right of 1.
The sketch
What earns the marks
Domain x ≠ 1; vertical asymptote x = 1 with f → −∞ from the left and +∞ from the right; slant asymptote y = x + 1 from the division f(x) = x + 1 + 1/(x − 1).
f′(x) = (x² − 2x)/(x − 1)² = x(x − 2)/(x − 1)²: increasing on (−∞, 0) and (2, ∞), decreasing on (0, 1) and (1, 2); local maximum (0, 0), local minimum (2, 4).
f″(x) = 2/(x − 1)³: concave down for x < 1, concave up for x > 1, with no inflection point (x = 1 is not in the domain).
A sketch showing both branches approaching the slant asymptote, the two extrema, and the correct behaviour at the vertical asymptote.
Check
f(−1) and f(0.5) are −0.5, below f(0) = 0, and f(1.5) and f(3) are 4.5, above f(2) = 4.
Where this goes wrong
Reporting a horizontal asymptote y = 1 from comparing leading coefficients. The numerator has higher degree, so the quotient grows without bound; dividing out gives the slant asymptote instead.
Leans on
Polynomial division, from Functions, graphs and inverses
Try a similar question
Let $f(x) = \dfrac{x^{2} + 1}{x - 1}$ . Find the $x$ -coordinate of its local minimum.
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Needs
The chain rule and logarithms
Compositions, nested rules, the natural logarithm and logarithmic differentiation.
30 questions
Needs
Limits
Limit laws, the indeterminate form 0/0, one-sided and infinite limits, limits at infinity and asymptotes.
30 questions

Key results

Worked example

Strategy

Working

Answer

Check

Where this goes wrong

On the paper

Leans on

Your turn

Practice questions

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Leans on

Try a similar question

Strategy

Working

Answer

Check

Where this goes wrong

Try a similar question

Strategy

Working

Answer

The sketch

What earns the marks

Check

Where this goes wrong

Leans on

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Related topics

The chain rule and logarithms

Limits