section 4.5 summary of curve sketchingms.mcmaster.ca/~rpshen/1za3/pptc4s5.pdfguidelines f. local...

Section 4.5 Summary of Curve Sketching

Ruipeng Shen

October 9-10th

Ruipeng Shen MATH 1ZA3 October 9-10th 1 / 21

Guidelines

The sketching should display the following aspects of the functions whenapplicable.

A. Domian

B. Intercepts Find the y -intercept by calculating f (0). Find the x-intercepts bysolving the equation f (x) = 0 when possible.

C. Symmetry Is the function odd, even or periodic?

D. Asymptotes The line y = L is a horizontal asymptote if

limx→+∞

f (x) = L or limx→−∞

f (x) = L.

The line x = a is a vertical asymptote if

limx→a−

f (x) = ±∞ or limx→a+

f (x) = ±∞.

E. Intervals of Increase or Decrease Calculate the first derivative f ′(x) anduse the I/D test.


Guidelines

F. Local Maximum and Minimum Values Find critical numbers and applythe first derivative test.

G. Concavity and Points of inflection Calculate f ′′(x) and use the concavitytest. (Optional) The derivatives at the inflection points are helpful.

Example

Sketch the curve y =2x2

x2 − 1.

Solution A. Domain = {x ∈ R|x 6= ±1}.B. The y -intercept is f (0) = 0. The x intercept is the solution to f (x) = 0,namely, x = 0.C. This function is even.

D. By the limit limx→±∞

2x2

x2 − 1= lim

x→±∞

2

1− 1/x2= 2+, We have a horizontal

asymptote y = 2 (↙↘).


Example 2x2

x2−1 (Continued)D (Continued). We consider the limits

limx→−1−

2x2

x2 − 1= +∞ 2

1+ − 1lim

x→−1+2x2

x2 − 1= −∞ 2

1− − 1

limx→1−

2x2

x2 − 1= −∞ 2

1− − 1lim

x→1+

2x2

x2 − 1= +∞ 2

1+ − 1

These give two vertical asymptotes x = −1 (↗↙) and x = 1 (↘↖).E-G. We can first calculate the derivatives

f (x) = 2 +2

x2 − 1= 2 + 2(x2 − 1)−1;

f ′(x) = 2 · [−(x2 − 1)−2] · 2x = −4x(x2 − 1)−2;

f ′′(x) = (−4x)′(x2 − 1)−2 + (−4x)[(x2 − 1)−2]′

= −4(x2 − 1)−2 + (−4x) · [−2(x2 − 1)−3] · 2x= (4− 4x2)(x2 − 1)−3 + 16x2(x2 − 1)−3

= (4 + 12x2)(x2 − 1)−3.


Example 2x2

x2−1 (Continued)

E-G (Continued). Now we can apply I/D Test and Concavity Test using

f ′(x) = −4x(x2 − 1)−2 f ′′(x) = (4 + 12x2)(x2 − 1)−3

Interval f ′(x) Monotonicity f ′′(x) Concavity

(−∞,−1) + Increasing + Upward

(−1, 0) + Increasing − Downward

(0, 1) − Decreasing − Downward

(1,∞) − Decreasing + Upward

The first derivative f ′(x) changes sign at x = 0, f (0) = 0 is a local maximumvalue.The second derivative f ′′(x) changes sign at x = ±1, but these are not in thedomain. Thus there are no point of inflection.


Sketching of y = 2x2

x2−1

Asymptotes: x=-1(↗↙) x=+1 (↘↖) y=2 (↙↘) -1 0 1

1

f(0)=0

Even


Example

Example

Sketch the curve y =x2√x + 1

.

Solution A. Domain = (−1,+∞).B. The y -intercept is f (0) = 0. The x intercept is the solution to f (x) = 0,namely, x = 0.C. This function is not odd, even or periodic.

D. By the limit limx→+∞

x2√x + 1

= +∞, there is no horizontal asymptote.

limx→−1+

x2√x + 1

= +∞ 1√0+

So we have a vertical asymptote x = −1 (↖).


Example f (x) = x2√x+1

(Continued)

E-G. We can first calculate the derivatives

f ′(x) =(x2)′

√x + 1− x2

(√x + 1

)′x + 1

=2x√x + 1− x2 · 1/(2

√x + 1)

x + 1

=4x(x + 1)− x2

2(x + 1)3/2=

3x2 + 4x

2(x + 1)3/23x2 + 4x = (3x + 4)x

f ′′(x) =(3x2 + 4x)′ · 2(x + 1)3/2 − (3x2 + 4x) · [2(x + 1)3/2]′

4(x + 1)3

=(6x + 4) · 2(x + 1)3/2 − (3x2 + 4x) · 2 · (3/2)(x + 1)1/2

4(x + 1)3

=(12x + 8)(x + 1)− 3 · (3x2 + 4x)

4(x + 1)5/2

=3x2 + 8x + 8

4(x + 1)5/2> 0 always concave upward

The function is decreasing in (−1, 0) and increasing in (0,∞). This is a singlelocal minimum f (0) = 0, no point of inflection.


Sketching of y = x2√x+1

Asymptotes: x=-1 (↖) -1 0

1

f(0)=0, f(3) = 9/2y

x


Example

Example

Sketch the curve y = xex .

Solution A. Domain = R.B. The y -intercept is f (0) = 0. The x intercept is the solution to f (x) = 0,namely, x = 0.C. This function is not odd, even or periodic.D. By the limit lim

x→+∞xex = +∞ and

limx→−∞

xex = limx→−∞

x

e−x= lim

x→−∞

1

−e−x= lim

x→−∞(−ex) = 0−

there is a horizontal asymptote y = 0 (↖).Since lim

x→axex = aea 6= ±∞, there is no vertical asymptote.


f ′(x) = (x)′ex + x(ex)′ = ex + xex = (x + 1)ex ;

f ′′(x) = (x + 1)′ex + (x + 1)(ex)′ = ex + (x + 1)ex = (x + 2)ex .


Example f (x) = xex (Continued)

We cut the domain into three intervals at x = −2,−1 (The roots of f ′(x) andf ′′(x))


(−∞,−2) − Decreasing − Downward

(−2,−1) − Decreasing + Upward

(−1,∞) + Increasing + Upward

The second derivative changes sign at x = −2 thus(−2, f (−2)) = (−2,−2e−2) is an inflection point.

The first derivative changes from negative to positive at x = −1. Thus thepoint (−1, f (−1)) = (−1,−1/e) is a local minimum.


Sketching of y = xex

-1f(-2)=-2e-2, f'(-2) = -e-2, f(-1) = -e-1, f(0)=0, f(1) = e

-2

1x

y


Example

Example

Sketch the curve y =cos x

2 + sin x.

Solution A. Domain = R.B. The y -intercept is f (0) = 1/2. The x intercept is the solution to f (x) = 0,namely, x = (k + 1/2)π for any integer k .C. This function is periodic f (x + 2π) = f (x).

D. The limits limx→±∞

cos x

2 + sin xdoes not exist. Thus there is no horizontal

asymptote.Since lim

x→af (x) = f (a) 6= ±∞, there is no vertical asymptote.


f ′(x) =(cos x)′(2 + sin x)− cos x(2 + sin x)′

(2 + sin x)2

=(− sin x)(2 + sin x) − cos x · cos x

(2 + sin x)2=−1− 2 sin x

(2 + sin x)2.


Example f (x) = cos x2+sin x (Continued)

We can continue to calculate the second derivative.

f ′(x) = −(1 + 2 sin x)(2 + sin x)−2

f ′′(x) = − d

dx

[(1 + 2 sin x)(2 + sin x)−2

]= −

{(1 + 2 sin x)′(2 + sin x)−2 + (1 + 2 sin x)

[(2 + sin x)−2

]′}= −

{2 cos x · (2 + sin x)−3+1 + (1 + 2 sin x)

[−2 · (2 + sin x)−3 · cos x

]′}= −(2 + sin x)−3 [2 cos x · (2 + sin x)− 2 cos x · (1 + 2 sin x)]

= −(2 + sin x)−3 · 2 cos x · (2 + sin x − 1− 2 sin x)

= −(2 + sin x)−3 · 2 cos x · (1− sin x)

=−2 cos x(1− sin x)

(2 + sin x)3.

Let us consider a few intervals cut from R: (−π/6, π/2), (π/2, 7π/6),(7π/6, 3π/2) and (3π/2, 11π/6).


Example f (x) = cos x2+sin x (Continued)

We have

f ′(x) =−(1 + 2 sin x)

(2 + sin x)2; f ′′(x) =

−2 cos x(1− sin x)

(2 + sin x)3.

Interval sin x f ′(x) Monotonicity f ′′(x) Concavity

(−π/6, π/2) > −1/2 − Decreasing − Downward

(π/2, 7π/6) > −1/2 − Decreasing + Upward

(7π/6, 3π/2) < −1/2 + Increasing + Upward

(3π/2, 11π/6) < −1/2 + Increasing − Downward

By the first derivative test, the point (7π/6, f (7π/6)) = (7π/6,−1/√

3) is alocal minimum; the point (−π/6, f (−π/6)) = (−π/6, 1/

√3) is a local

maximum.

The points (π/2, f (π/2)) = (π/2, 0) and (3π/2, f (3π/2)) = (3π/2, 0) arepoints of inflection, since f ′′(x) changes sign at these points;


Sketching of y = cos x2+sin x

-𝛑 𝛑 7𝛑 3𝛑 11𝛑 6 2 6 2 6

𝛑 2𝛑-2𝛑 -𝛑

f(-𝛑/6)=1/√3 f(𝛑/2)=0 f(7𝛑/6)=-1/√3 f(3𝛑/2)=0f(11𝛑/6)=1/√3 f(0)=1/2

y

x

1

2𝛑-periodic


Slant Asymptote

DefinitionThe line y = mx + b is called a slant asymptote if

limx→∞

[f (x)− (mx + b)] = 0

Remark We have m = limx→∞

f (x)

xand b = lim

x→∞[f (x)−mx ].

f(x)-(mx+b)

y=f(x)

y=mx+b

y

x

Figure : The distance between the curve y = f (x) and the line y = mx + b approaches 0.


Example

Example

Sketch the curve y =x3

x2 + 3.

Solution A. Domain = R.B. The y -intercept is f (0) = 0. The x intercept is the solution to f (x) = 0,namely, x = 0.C. This function is odd.D. Since lim

x→af (x) = f (a) 6= ±∞, there is no vertical asymptote.

The limit m = limx→±∞

f (x)

x= lim

x→±∞

x2

x2 + 3= lim

x→±∞

1

1 + 3/x2= 1 exists.

In addition, we have

b = limx→±∞

[f (x)− 1 · x ] = limx→±∞

x3 − (x2 + 3)x

x2 + 3= lim

x→±∞

−3x

x2 + 3= 0.

As a result, the line y = x is a slant asymptote (↙↗).


Example f (x) = x3

x2+3


f ′(x) =(x3)′(x2 + 3)− x3 · (x2 + 3)′

(x2 + 3)2=

3x2(x2 + 3)− x3 · 2x(x2 + 3)2

=x4 + 9x2

(x2 + 3)2= 1 + 3 · x2 − 3

(x2 + 3)2

f ′′(x) = 3 · ddx

[(x2 − 3)(x2 + 3)−2

]= 3 ·

{(x2 − 3)′(x2 + 3)−2 + (x2 − 3)[(x2 + 3)−2]′

}= 3

{2x · (x2 + 3)−3+1 + (x2 − 3) · (−2)(x2 + 3)−3 · 2x

}= 3 · 2x(x2 + 3)−3 ·

[(x2 + 3)− 2(x2 − 3)

]=

6x(9− x2)

(x2 + 3)3.

We break the domain into 4 intervals at the points x = −3, 0, 3.


Example f (x) = x3

x2+3

We have

f ′(x) =x2(x2 + 9)

(x2 + 3)2f ′′(x) =

6x(9− x2)

(x2 + 3)3


(−∞,−3) + Increasing + Upward

(−3, 0) + Increasing − Downward

(0, 3) + Increasing + Upward

(3,∞) + Increasing − Downward

There are no local extreme values. But we have three points of inflection

(−3, f (−3)) = (−3,−9/4), f ′(−3) = 9/8;

(0, f (0)) = (0, 0), f ′(0) = 0;

(3, f (3)) = (3, 9/4), f ′(3) = 9/8.


Sketching of y = x3

x2+3

Asymptotes: y=x (↙↗)Odd Function

1 2 3

-3 0 3f(0)=0 f'(0)=0f(-3)=-9/4 f'(-3)=9/8f(3)=9/4 f'(3)=9/8

y

x

y=x


section 4.5 summary of curve sketchingms.mcmaster.ca/~rpshen/1za3/pptc4s5.pdfguidelines f. local...

Documents