calculus curve sketching (i) - la citadelle · calculus curve sketching (i) 1. sketch the grapf of...

CALCULUS Curve Sketching (I)

1. Sketch the grapf of the following polynomial function: f(x) =(x + 3)(x + 1)(x− 3)

−4

2. Sketch the grapf of the following polynomial function: f(x) =(x + 3)(x− 1)(x− 2)

−4


−4


2


−2


3


3


3


−5

10. Sketch the grapf of the following polynomial function: f(x) =(x + 3)(x + 2)(x + 1)

−5

c© 2009 La Citadelle 1 of 21 www.la-citadelle.com


Solutions:

1. f(x) =(x + 3)(x + 1)(x− 3)

−4

By expanding, f(x) =−9− 9x + x2 + x3

−4DomainThe function f(x) is a polynomial function. Therefore the domain is Df = R .

Symmetry

f(−x) =−9− 9(−x) + (−x)2 + (−x)3

−4=−9 + 9x + x2 − x3

−4f(−x) 6= f(x) f(−x) 6= −f(x)

Therefore the function f(x) is neither even nor odd function.

ZerosThe zeros of the function f(x) are: x1 = −3 x2 = −1 x3 = 3

y-intercept

y − int = f(0) =−9−4

= 2.250

AsymptotesThe function f(x) is a polynomial function of degree 3. Therefore the function does not have any kind ofasymptotes.

Critical Numbers

f ′(x) =−9 + 2x + 3x2

−4

Critical numbers are the solutions of the equation f ′(x) = 0 or −9 + 2x + 3x2 = 0

x =−2±

√(−2)2 − 4(3)(−9)

6

x4 =−2−

√(112)

6= −2.097 y4 = f(x4) = −1.262

x5 =−2 +

√(112)

6= 1.431 y5 = f(x5) = 4.225

Sign Chart for the First Derivative f ′(x)

x −2.097 1.431

f(x) ↓ −1.262 ↑ 4.225 ↓

f ′(x) − 0 + 0 −

Increasing and Decreasing Intervals

The function f(x) is decreasing over (−∞,−2.097) and over (1.431,∞) and is increasing over (−2.097, 1.431).

Maximum and Minimum Points

The function f(x) has a minimum point at P4(−2.097,−1.262) and a maximum point at P5(1.431, 4.225).

Concavity Intervals

The second derivative of the function f(x) is given by: f ′′(x) =2 + 6x

−4



The second derivative f ′′(x) is zero when: f ′′(x) = 0 2 + 6x = 0 x6 =−13

y6 = f(x6) = 1.481

Sign Chart for the Second Derivative f ′′(x)

x −0.333

f(x) ^ 1.481 _

f ′′(x) + 0 −

Inflection PointsThere is an inflection point at P6 = (−0.333, 1.481)

Graph

- x

6y

−4 −2 2 4

−4

−2

2

4

(−2.10,−1.26)

(1.43, 4.23)

(−0.33, 1.48)



2. f(x) =(x + 3)(x− 1)(x− 2)

−4

By expanding, f(x) =6− 7x + x3


Symmetry

f(−x) =6− 7(−x) + (−x)3

−4=

6 + 7x− x3

−4f(−x) 6= f(x) f(−x) 6= −f(x)


ZerosThe zeros of the function f(x) are: x1 = −3 x2 = 1 x3 = 2

y-intercept

y − int = f(0) =6−4

= −1.500


Critical Numbers

f ′(x) =−7 + 3x2

−4

Critical numbers are the solutions of the equation f ′(x) = 0 or −7 + 3x2 = 0

x =0±

√(0)2 − 4(3)(−7)

6

x4 =0−

√(84)

6= −1.528 y4 = f(x4) = −3.282

x5 =0 +

√(84)

6= 1.528 y5 = f(x5) = 0.282


x −1.528 1.528

f(x) ↓ −3.282 ↑ 0.282 ↓

f ′(x) − 0 + 0 −





Concavity Intervals

The second derivative of the function f(x) is given by: f ′′(x) =6x

−4The second derivative f ′′(x) is zero when: f ′′(x) = 0 6x = 0 x6 = 0 y6 = f(x6) = −1.500




x 0.000

f(x) ^ −1.500 _

f ′′(x) + 0 −

Inflection PointsThere is an inflection point at P6 = (0.000,−1.500)

Graph

- x

6y

−4 −2 2 4

−4

−2

2

4

(−1.53,−3.28)

(1.53, 0.28)

(0.00,−1.50)



3. f(x) =(x + 1)(x− 0)(x− 2)

−4

By expanding, f(x) =−2x− x2 + x3


Symmetry

f(−x) =−2(−x)− (−x)2 + (−x)3

−4=

2x− x2 − x3

−4f(−x) 6= f(x) f(−x) 6= −f(x)



y-intercept

y − int = f(0) =0−4

= 0.000


Critical Numbers

f ′(x) =−2− 2x + 3x2

−4

Critical numbers are the solutions of the equation f ′(x) = 0 or −2− 2x + 3x2 = 0

x =2±

√(2)2 − 4(3)(−2)

6

x4 =2−

√(28)

6= −0.549 y4 = f(x4) = −0.158

x5 =2 +

√(28)

6= 1.215 y5 = f(x5) = 0.528


x −0.549 1.215

f(x) ↓ −0.158 ↑ 0.528 ↓

f ′(x) − 0 + 0 −





Concavity Intervals

The second derivative of the function f(x) is given by: f ′′(x) =−2 + 6x

−4



The second derivative f ′′(x) is zero when: f ′′(x) = 0 − 2 + 6x = 0 x6 =13

y6 = f(x6) = 0.185


x 0.333

f(x) ^ 0.185 _

f ′′(x) + 0 −

Inflection PointsThere is an inflection point at P6 = (0.333, 0.185)

Graph

- x

6y

−4 −2 2 4

−4

−2

2

4

(−0.55,−0.16)

(1.22, 0.53)

(0.33, 0.19)



4. f(x) =(x + 3)(x + 2)(x− 2)

2

By expanding, f(x) =−12− 4x + 3x2 + x3

2DomainThe function f(x) is a polynomial function. Therefore the domain is Df = R .

Symmetry

f(−x) =−12− 4(−x) + 3(−x)2 + (−x)3

2=−12 + 4x + 3x2 − x3

2f(−x) 6= f(x) f(−x) 6= −f(x)



y-intercept

y − int = f(0) =−12

2= −6.000


Critical Numbers

f ′(x) =−4 + 6x + 3x2

2Critical numbers are the solutions of the equation f ′(x) = 0 or −4 + 6x + 3x2 = 0

x =−6±

√(−6)2 − 4(3)(−4)

6

x4 =−6−

√(84)

6= −2.528 y4 = f(x4) = 0.564

x5 =−6 +

√(84)

6= 0.528 y5 = f(x5) = −6.564


x −2.528 0.528

f(x) ↑ 0.564 ↓ −6.564 ↑

f ′(x) + 0 − 0 +


The function f(x) is increasing over (−∞,−2.528) and over (0.528,∞) and is decreasing over (−2.528, 0.528).


The function f(x) has a maximum point at P4(−2.528, 0.564) and a minimum point at P5(0.528,−6.564).

Concavity Intervals


2The second derivative f ′′(x) is zero when: f ′′(x) = 0 6 + 6x = 0 x6 = −1 y6 = f(x6) = −3.000




x −1.000

f(x) _ −3.000 ^

f ′′(x) − 0 +

Inflection PointsThere is an inflection point at P6 = (−1.000,−3.000)

Graph

- x

6y

−4 −2 2 4

−4

−2

2

4

(−2.53, 0.56)

(0.53,−6.56)

(−1.00,−3.00)



5. f(x) =(x + 1)(x− 2)(x− 3)

−2

By expanding, f(x) =6 + x− 4x2 + x3


Symmetry

f(−x) =6 + (−x)− 4(−x)2 + (−x)3

−2=

6− x− 4x2 − x3

−2f(−x) 6= f(x) f(−x) 6= −f(x)



y-intercept

y − int = f(0) =6−2

= −3.000


Critical Numbers

f ′(x) =1− 8x + 3x2

−2

Critical numbers are the solutions of the equation f ′(x) = 0 or 1− 8x + 3x2 = 0

x =8±

√(8)2 − 4(3)(1)

6

x4 =8−

√(52)

6= 0.131 y4 = f(x4) = −3.032

x5 =8 +

√(52)

6= 2.535 y5 = f(x5) = 0.440


x 0.131 2.535

f(x) ↓ −3.032 ↑ 0.440 ↓

f ′(x) − 0 + 0 −


The function f(x) is decreasing over (−∞, 0.131) and over (2.535,∞) and is increasing over (0.131, 2.535).


The function f(x) has a minimum point at P4(0.131,−3.032) and a maximum point at P5(2.535, 0.440).

Concavity Intervals


−2




y6 = f(x6) = −1.296


x 1.333

f(x) ^ −1.296 _

f ′′(x) + 0 −


Graph

- x

6y

−4 −2 2 4

−4

−2

2

4

(0.13,−3.03)

(2.54, 0.44)

(1.33,−1.30)



6. f(x) =(x + 3)(x + 1)(x− 3)

3

By expanding, f(x) =−9− 9x + x2 + x3


Symmetry

f(−x) =−9− 9(−x) + (−x)2 + (−x)3

3=−9 + 9x + x2 − x3

3f(−x) 6= f(x) f(−x) 6= −f(x)



y-intercept

y − int = f(0) =−93

= −3.000


Critical Numbers

f ′(x) =−9 + 2x + 3x2


x =−2±

√(−2)2 − 4(3)(−9)

6

x4 =−2−

√(112)

6= −2.097 y4 = f(x4) = 1.683

x5 =−2 +

√(112)

6= 1.431 y5 = f(x5) = −5.634


x −2.097 1.431

f(x) ↑ 1.683 ↓ −5.634 ↑

f ′(x) + 0 − 0 +





Concavity Intervals


3


y6 = f(x6) = −1.975




x −0.333

f(x) _ −1.975 ^

f ′′(x) − 0 +

Inflection PointsThere is an inflection point at P6 = (−0.333,−1.975)

Graph

- x

6y

−4 −2 2 4

−4

−2

2

4

(−2.10, 1.68)

(1.43,−5.63)

(−0.33,−1.98)



7. f(x) =(x + 2)(x− 0)(x− 1)

3

By expanding, f(x) =−2x + x2 + x3


Symmetry

f(−x) =−2(−x) + (−x)2 + (−x)3

3=

2x + x2 − x3

3f(−x) 6= f(x) f(−x) 6= −f(x)



y-intercept

y − int = f(0) =03

= 0.000


Critical Numbers

f ′(x) =−2 + 2x + 3x2


x =−2±

√(−2)2 − 4(3)(−2)

6

x4 =−2−

√(28)

6= −1.215 y4 = f(x4) = 0.704

x5 =−2 +

√(28)

6= 0.549 y5 = f(x5) = −0.210


x −1.215 0.549

f(x) ↑ 0.704 ↓ −0.210 ↑

f ′(x) + 0 − 0 +





Concavity Intervals


3


y6 = f(x6) = 0.247




x −0.333

f(x) _ 0.247 ^

f ′′(x) − 0 +


Graph

- x

6y

−4 −2 2 4

−4

−2

2

4

(−1.22, 0.70)

(0.55,−0.21)

(−0.33, 0.25)



8. f(x) =(x + 1)(x− 1)(x− 3)

3

By expanding, f(x) =3− x− 3x2 + x3


Symmetry

f(−x) =3− (−x)− 3(−x)2 + (−x)3

3=

3 + x− 3x2 − x3

3f(−x) 6= f(x) f(−x) 6= −f(x)



y-intercept

y − int = f(0) =33

= 1.000


Critical Numbers

f ′(x) =−1− 6x + 3x2

3Critical numbers are the solutions of the equation f ′(x) = 0 or −1− 6x + 3x2 = 0

x =6±

√(6)2 − 4(3)(−1)

6

x4 =6−

√(48)

6= −0.155 y4 = f(x4) = 1.026

x5 =6 +

√(48)

6= 2.155 y5 = f(x5) = −1.026


x −0.155 2.155

f(x) ↑ 1.026 ↓ −1.026 ↑

f ′(x) + 0 − 0 +





Concavity Intervals


3The second derivative f ′′(x) is zero when: f ′′(x) = 0 − 6 + 6x = 0 x6 = 1 y6 = f(x6) = 0.000




x 1.000

f(x) _ 0.000 ^

f ′′(x) − 0 +

Inflection PointsThere is an inflection point at P6 = (1.000, 0.000)

Graph

- x

6y

−4 −2 2 4

−4

−2

2

4

(−0.15, 1.03)

(2.15,−1.03)

(1.00, 0.00)



9. f(x) =(x + 1)(x− 2)(x− 3)

−5

By expanding, f(x) =6 + x− 4x2 + x3


Symmetry

f(−x) =6 + (−x)− 4(−x)2 + (−x)3

−5=

6− x− 4x2 − x3

−5f(−x) 6= f(x) f(−x) 6= −f(x)



y-intercept

y − int = f(0) =6−5

= −1.200


Critical Numbers

f ′(x) =1− 8x + 3x2

−5

Critical numbers are the solutions of the equation f ′(x) = 0 or 1− 8x + 3x2 = 0

x =8±

√(8)2 − 4(3)(1)

6

x4 =8−

√(52)

6= 0.131 y4 = f(x4) = −1.213

x5 =8 +

√(52)

6= 2.535 y5 = f(x5) = 0.176


x 0.131 2.535

f(x) ↓ −1.213 ↑ 0.176 ↓

f ′(x) − 0 + 0 −


The function f(x) is decreasing over (−∞, 0.131) and over (2.535,∞) and is increasing over (0.131, 2.535).


The function f(x) has a minimum point at P4(0.131,−1.213) and a maximum point at P5(2.535, 0.176).

Concavity Intervals


−5




y6 = f(x6) = −0.519


x 1.333

f(x) ^ −0.519 _

f ′′(x) + 0 −


Graph

- x

6y

−4 −2 2 4

−4

−2

2

4

(0.13,−1.21)

(2.54, 0.18)

(1.33,−0.52)



10. f(x) =(x + 3)(x + 2)(x + 1)

−5

By expanding, f(x) =6 + 11x + 6x2 + x3


Symmetry

f(−x) =6 + 11(−x) + 6(−x)2 + (−x)3

−5=

6− 11x + 6x2 − x3

−5f(−x) 6= f(x) f(−x) 6= −f(x)


ZerosThe zeros of the function f(x) are: x1 = −3 x2 = −2 x3 = −1

y-intercept

y − int = f(0) =6−5

= −1.200


Critical Numbers

f ′(x) =11 + 12x + 3x2

−5

Critical numbers are the solutions of the equation f ′(x) = 0 or 11 + 12x + 3x2 = 0

x =−12±

√(−12)2 − 4(3)(11)

6

x4 =−12−

√(12)

6= −2.577 y4 = f(x4) = −0.077

x5 =−12 +

√(12)

6= −1.423 y5 = f(x5) = 0.077


x −2.577 −1.423

f(x) ↓ −0.077 ↑ 0.077 ↓

f ′(x) − 0 + 0 −


The function f(x) is decreasing over (−∞,−2.577) and over (−1.423,∞) and is increasing over (−2.577,−1.423).


The function f(x) has a minimum point at P4(−2.577,−0.077) and a maximum point at P5(−1.423, 0.077).

Concavity Intervals


−5The second derivative f ′′(x) is zero when: f ′′(x) = 0 12 + 6x = 0 x6 = −2 y6 = f(x6) = 0.000




x −2.000

f(x) ^ 0.000 _

f ′′(x) + 0 −


Graph

- x

6y

−4 −2 2 4

−4

−2

2

4

(−2.58,−0.08)

(−1.42, 0.08)(−2.00, 0.00)


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