Curve Sketching

The graph of a function provides invaluable information about its behavior. An 8-step strategy using derivatives, domain, symmetry, asymptotes, intercepts, intervals of increase/decrease, local extrema, concavity, and final sketch, organizes all the information gathered from calculus into a coherent picture.

The 8-Step Strategy

Step 1: Domain. Determine the domain of the function (the set of $x$ values for which $f (x)$ is defined) and any points of discontinuity.

Step 2: Symmetry. Determine if the function is even, odd, or neither. Also, check if the function is periodic.

(a) Even functions. The function is even if $f (- x) = f (x)$ for all $x$ in the domain. If $f$ is even, plot $y = f (x)$ for $x > 0$ and reflect about the $y$ -axis.

Graph of an even function symmetric about the y-axis

(b) Odd functions. The function is odd if $f (- x) = - f (x)$ for all $x$ in the domain. If $f$ is odd, plot $y = f (x)$ for $x > 0$ and rotate $180°$ about the origin.

Graph of an odd function symmetric about the origin

(c) Periodic functions. The function is periodic with period $T > 0$ if $f (x + T) = f (x)$ for all $x$ . Plot $y = f (x)$ for one period and repeat.

Graph of a periodic function showing one period and its repetition

Step 3: Asymptotes (if any). Determine the vertical, horizontal, and oblique asymptotes of the function.

Step 4: Intercepts.

(a) The $y$ -intercept: substitute $x = 0$ into $f (x)$ .

(b) The $x$ -intercepts: solve $f (x) = 0$ . (Skip if difficult.)

Step 5: Intervals of increase or decrease. Calculate f'(x) and determine its sign. Where f'>0, the function is increasing; where f'<0, it is decreasing.

Step 6: Local maxima and minima. Find the critical points (f'(x_0)=0 or f'(x_0) does not exist). Use the First or Second Derivative Test to classify each.

Step 7: Concavity and inflection points. Calculate f''(x) and determine its sign. If f''(x)>0, the function is concave up; if f''(x)<0, it is concave down. Where the direction of concavity changes, there is an inflection point.

Step 8: Sketch the curve. Draw asymptotes, plot important points (intercepts, local extrema, inflection points), and draw a smooth curve using all information gathered.

Examples

Sketch the graph of

f (x) = \frac{2 x - 1}{3 x - 6} .

Solution

**1. Domain:**

{x ∣ 3 x - 6 \neq 0} = {x ∣ x \neq 2} = (- \infty, 2) \cup (2, + \infty) .

**2. Symmetry:** None, since

f (- x) = \frac{- 2 x - 1}{- 3 x - 6} \neq f (x) or - f (x) .

**3. Asymptotes.** (a) **Horizontal asymptote:**

lim_{x \to \pm \infty} \frac{2 x - 1}{3 x - 6} = lim_{x \to \pm \infty} \frac{2 x}{3 x} = \frac{2}{3} .

The line

y = 2 / 3

is the horizontal asymptote; there is no oblique asymptote. (b) **Vertical asymptote:**

lim_{x \to 2^{+}} \frac{2 x - 1}{3 x - 6} \overset{[\frac{3}{0^{+}}]}{=} + \infty, lim_{x \to 2^{-}} \frac{2 x - 1}{3 x - 6} \overset{[\frac{3}{0^{-}}]}{=} - \infty .

The line

x = 2

is a vertical asymptote. **4. Intercepts.** (a)

y

-intercept:

f (0) = \frac{- 1}{- 6} = \frac{1}{6}

, so the graph meets the

y

-axis at

(0, 1 / 6)

. (b)

x

-intercept:

\frac{2 x - 1}{3 x - 6} = 0 \Rightarrow x = \frac{1}{2}

, so the graph meets the

x

-axis at

(1 / 2, 0)

. So far the graph looks like this:

Partial sketch of y = (2x-1)/(3x-6) showing asymptotes and intercepts

**5. Intervals of increase or decrease:** f'(x)=\frac{2(3x-6)-3(2x-1)}{(3x-6)^{2}}=\frac{-9}{(3x-6)^{2}}<0. Because f'(x)<0 everywhere but

f

is discontinuous at

x = 2

f

is decreasing on

(- \infty, 2)

and on

(2, + \infty)

. **6. Local maxima and minima:** There are no critical points since f'(x)\neq0. Although f'(2) does not exist,

x = 2

is not a critical point because it is not in the domain of

f

. **7. Concavity and inflection points:** f''(x)=\frac{54}{(3x-6)^{3}}. f''(x)>0\Longleftrightarrow x>2,\qquad f''(x)<0\Longleftrightarrow x<2.

f

is concave up on

(2, + \infty)

and concave down on

(- \infty, 2)

. Although the sign of f'' changes at

x = 2

, there is no inflection point there because

x = 2

is not in the domain. **8. Sketch:**

Complete graph of y = (2x-1)/(3x-6) — Graph of $y = \frac{2 x - 1}{3 x - 6}$ .

Sketch the graph of

f (x) = 5 x + 3 \sqrt{x^{2} - 1} .

Solution

**1. Domain:**

Dom (f) = {x ∣ x^{2} - 1 \geq 0} = (- \infty, - 1] \cup [1, + \infty) .

**2. Symmetry:** None. **3. Asymptotes:** No vertical asymptote (continuous on its domain). Since

lim_{x \to + \infty} f (x) = + \infty

, there is no horizontal asymptote as

x \to + \infty

. As

x \to - \infty

, the limit is indeterminate

(- \infty + \infty)

; multiplying by the conjugate gives

lim_{x \to - \infty} f (x) = - \infty

. No horizontal asymptote. **Oblique asymptote as

x \to + \infty

:**

lim_{x \to + \infty} \frac{f (x)}{x} = lim_{x \to + \infty} (5 + 3 \sqrt{1 - \frac{1}{x^{2}}}) = 8.

lim_{x \to + \infty} (f (x) - 8 x) = lim_{x \to + \infty} (- 3 x + 3 \sqrt{x^{2} - 1}) = 0.

y = 8 x

is an oblique asymptote as

x \to + \infty

. **Oblique asymptote as

x \to - \infty

:**

lim_{x \to - \infty} \frac{f (x)}{x} = lim_{x \to - \infty} (5 - 3 \sqrt{1 - \frac{1}{x^{2}}}) = 2.

lim_{x \to - \infty} (f (x) - 2 x) = 0.

y = 2 x

is an oblique asymptote as

x \to - \infty

. **4. Intercepts:** Since

x = 0

is not in the domain, there is no

y

-intercept. Setting

f (x) = 0

leads to

16 x^{2} = - 9

, which has no real solution, so there are no

x

-intercepts. **5. Intervals of increase or decrease:** f'(x)=5+\frac{3x}{\sqrt{x^{2}-1}}. Setting f'(x)=0: squaring after rearranging gives

x = \pm 5 / 4

. Only

x = - 5 / 4

is valid (the equation requires

3 x \leq 0

). The sign diagram:

Sign diagram for f'(x) = 5 + 3x/sqrt(x^2 - 1)

**6. Local maxima and minima:** The critical point is

x = - 5 / 4

, where f' changes from

+

-

, so

f (- 5 / 4) = - 4

is a **local minimum**. **7. Concavity and inflection points:** f''(x)=-\frac{3}{(x^{2}-1)^{3/2}}<0 for all

x

in the domain. The graph is concave down everywhere; no inflection points. **8. Sketch:**

Graph of y = 5x + 3*sqrt(x^2 - 1) with oblique asymptotes y = 8x and y = 2x — Graph of $y = 5 x + 3 \sqrt{x^{2} - 1}$ and its oblique asymptotes.

Sketch the graph of

f (x) = \frac{x}{\sqrt[3]{x^{2} - 1}} .

Solution

**1. Domain:** All

x

except

x = \pm 1

(where the denominator is zero). **2. Symmetry:** The function is odd:

f (- x) = \frac{- x}{\sqrt[3]{(- x)^{2} - 1}} = - f (x) .

The graph is symmetric about the origin, so we may sketch for

x \geq 0

and use symmetry. **3. Asymptotes.** (a) **Vertical asymptotes:** At

x = 1

(by symmetry, also at

x = - 1

lim_{x \to 1^{+}} \frac{x}{\sqrt[3]{x^{2} - 1}} \overset{[\frac{1}{0^{+}}]}{=} + \infty, lim_{x \to 1^{-}} \frac{x}{\sqrt[3]{x^{2} - 1}} \overset{[\frac{1}{0^{-}}]}{=} - \infty .

(b) **Horizontal asymptote:**

lim_{x \to + \infty} f (x) = lim_{x \to + \infty} x^{1 / 3} = + \infty

. No horizontal asymptote. (c) **Oblique asymptote:**

m = lim_{x \to + \infty} f (x) / x = lim_{x \to + \infty} 1 / \sqrt[3]{x^{2} - 1} = 0

, but

lim_{x \to + \infty} f (x) = + \infty

. No oblique asymptote. **4. Intercepts:** Both

x

- and

y

-intercepts are $0$. **5. Intervals of increase or decrease:** f'(x)=\frac{x^{2}-3}{3(x^{2}-1)^{4/3}}. The denominator is always nonnegative, so the sign of f' is determined by

x^{2} - 3 = (x - \sqrt{3}) (x + \sqrt{3})

. f'<0 between

- \sqrt{3}

and

\sqrt{3}

; f'>0 elsewhere.

Sign diagram for f'(x) = (x^2 - 3)/(3(x^2-1)^(4/3))

**6. Local maxima and minima:** For

x \geq 0

f (\sqrt{3}) = \sqrt{3} / \sqrt[3]{2} \approx 1.375

is a **local minimum**. The graph for

x \geq 0

looks like this:

Partial sketch of y = x/cbrt(x^2-1) for x loading=

**7. Concavity and inflection points:** f''(x)=\frac{2x(9-x^{2})}{9(x^{2}-1)^{7/3}}.

Sign diagram for f''(x) showing concavity regions

x = 3

, the direction of concavity changes and

f

is continuous, so

(3, f (3)) = (3, 1.5)

is an **inflection point**. (At

x = 1

, concavity also changes direction, but there is no inflection point because

f

is discontinuous there.) **8. Sketch:** Sketch for

x \geq 0

using the information above, then apply the odd symmetry to obtain the full graph.

Complete graph of y = x/cbrt(x^2 - 1) — Graph of $y = \frac{x}{\sqrt[3]{x^{2} - 1}}$ .

Sketch the graph of

f (x) = \frac{\sin^{3} x}{1 + \cos x} .

Hint: First simplify and show that $f (x) = \sin x - \frac{1}{2} \sin 2 x$ provided $1 + \cos x \neq 0$ .

Solution

**Simplification.** Using

\sin^{2} x = 1 - \cos^{2} x = (1 - \cos x) (1 + \cos x)

f (x) = \frac{\sin x \cdot \sin^{2} x}{1 + \cos x} = \frac{\sin x (1 - \cos x) (1 + \cos x)}{1 + \cos x} = \sin x (1 - \cos x) = \sin x - \frac{1}{2} \sin 2 x

provided

1 + \cos x \neq 0

. **1. Domain:**

f

is defined for all

x

such that

\cos x \neq - 1

, i.e.,

Dom (f) = {x ∣ x \neq (2 k + 1) π, k \in ℤ} .

**2. Symmetry:** The function is odd (

f (- x) = - f (x)

) and periodic with period

2 π

. We sketch

f

[0, π)

and use both symmetries for the full graph. Note that

f

has a removable discontinuity at

x = π

lim_{x \to π} f (x) = \sin π - \frac{1}{2} \sin 2 π = 0.

**3. Asymptotes:** None (continuous and periodic). **4. Intercepts:**

y

-intercept:

f (0) = 0

. Setting

\sin x (1 - \cos x) = 0

[0, π)

gives only

x = 0

. The graph touches the

x

-axis only at the origin (with

(π, 0)

a hole). **5. Intervals of increase or decrease:** f'(x)=\cos x-\cos2x. Setting f'(x)=0: substituting

\cos 2 x = 2 \cos^{2} x - 1

gives

2 \cos^{2} x - \cos x - 1 = 0

, so

\cos x = 1 or \cos x = - \frac{1}{2} .

[0, π)

x = 0

(from

\cos x = 1

) and

x = 2 π / 3

(from

\cos x = - 1 / 2

). Testing: f'(\pi/2)=0-(-1)>0,\qquad f'(5\pi/6)=-\tfrac{\sqrt{3}}{2}-\tfrac{1}{2}<0.

Sign diagram for f'(x) = cos x - cos 2x on [0, pi]

**6. Local maxima and minima:** In

[0, π]

f (2 π / 3) = \sin \frac{2 π}{3} - \frac{1}{2} \sin \frac{4 π}{3} = \frac{\sqrt{3}}{2} + \frac{\sqrt{3}}{4} = \frac{3 \sqrt{3}}{4}

is a **local maximum**. **7. Concavity and inflection points:** f''(x)=-\sin x+2\sin2x=\sin x(4\cos x-1). Setting f''(x)=0 in

[0, π]

\sin x = 0

gives

x = 0, π

;

\cos x = 1 / 4

gives

x = \arccos (1 / 4) \approx 1.318

. Testing: f''(\pi/6)>0 and f''(\pi/2)<0, so

f

has an **inflection point** at

x = \arccos (1 / 4)

Sign diagram for f''(x) showing inflection at arccos(1/4)

**8. Sketch:** First sketch

f

[0, π)

and remove the point

(π, 0)

Graph of f for 0 <= x < pi — Graph of $f$ for $0 \leq x < π$ .

Then apply odd symmetry to get the graph on

(- π, π)

Graph of f for -pi < x < pi — Graph of $f$ for $- π < x < π$ .

Finally, extend by periodicity to obtain the complete graph:

Complete graph of f(x) = sin^3(x)/(1 + cos x) — Graph of $f (x) = \frac{\sin^{3} x}{1 + \cos x}$ .

Frequently Asked Questions

Do I need to perform all 8 steps for every curve?

Not necessarily. For simpler functions, some steps may add little information (e.g., a polynomial has no asymptotes). However, working through all steps is good practice and ensures nothing is missed. For a rational function or a function involving radicals or trigonometry, each step typically reveals something useful.

In what order should I use the 8 steps?

The order given is recommended because early steps (domain, symmetry, asymptotes) constrain the overall shape, while later steps (derivatives) fill in the details. However, you may reorder if it suits the function, for example, computing f' early can help confirm asymptotic behavior.

What if the second derivative is too complicated to compute?

Use the First Derivative Test exclusively. Concavity information is useful but not always essential for a good sketch. If you can determine the intervals of increase/decrease and the local extrema accurately, the overall shape of the graph will be correct.

The 8-Step Strategy

Examples

Frequently Asked Questions

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