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.
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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 (↙↘).
Ruipeng Shen MATH 1ZA3 October 9-10th 3 / 21
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.
Ruipeng Shen MATH 1ZA3 October 9-10th 4 / 21
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.
Ruipeng Shen MATH 1ZA3 October 9-10th 5 / 21
Sketching of y = 2x2
x2−1
Asymptotes: x=-1(↗↙) x=+1 (↘↖) y=2 (↙↘) -1 0 1
1
f(0)=0
Even
Ruipeng Shen MATH 1ZA3 October 9-10th 6 / 21
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 (↖).
Ruipeng Shen MATH 1ZA3 October 9-10th 7 / 21
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.
Ruipeng Shen MATH 1ZA3 October 9-10th 8 / 21
Sketching of y = x2√x+1
Asymptotes: x=-1 (↖) -1 0
1
f(0)=0, f(3) = 9/2y
x
Ruipeng Shen MATH 1ZA3 October 9-10th 9 / 21
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.
E-G. We can first calculate the derivatives
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 .
Ruipeng Shen MATH 1ZA3 October 9-10th 10 / 21
Example f (x) = xex (Continued)
We cut the domain into three intervals at x = −2,−1 (The roots of f ′(x) andf ′′(x))
Interval f ′(x) Monotonicity f ′′(x) Concavity
(−∞,−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.
Ruipeng Shen MATH 1ZA3 October 9-10th 11 / 21
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
Ruipeng Shen MATH 1ZA3 October 9-10th 12 / 21
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.
E-G. We can first calculate the derivatives
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.
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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).
Ruipeng Shen MATH 1ZA3 October 9-10th 14 / 21
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;
Ruipeng Shen MATH 1ZA3 October 9-10th 15 / 21
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
Ruipeng Shen MATH 1ZA3 October 9-10th 16 / 21
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.
Ruipeng Shen MATH 1ZA3 October 9-10th 17 / 21
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 (↙↗).
Ruipeng Shen MATH 1ZA3 October 9-10th 18 / 21
Example f (x) = x3
x2+3
E-G. We can first calculate the derivatives
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.
Ruipeng Shen MATH 1ZA3 October 9-10th 19 / 21
Example f (x) = x3
x2+3
We have
f ′(x) =x2(x2 + 9)
(x2 + 3)2f ′′(x) =
6x(9− x2)
(x2 + 3)3
Interval f ′(x) Monotonicity f ′′(x) Concavity
(−∞,−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.
Ruipeng Shen MATH 1ZA3 October 9-10th 20 / 21
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
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