MCV4U Unit 3 Curve Sketching PDF

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MCV4U Unit 3: CurveSketching1a. Given:𝑓(𝑥) = 𝑥 2 + 2𝑥 − 15To find the x-intercepts, I set y=0 and solved for x:𝑓(𝑥) = 𝑥 2 + 2𝑥 − 15 𝑓(𝑥) = (𝑥 + 5)(𝑥 − 3) 𝑥 + 5 = 0 𝑥 =− 5 𝑥 − 3 = 0 𝑥 = 3To find the y-intercepts, I set x=0 and solved for y:𝑓(0) = 0 2 + 2(0) − 15 𝑓(0) =− 15 𝑦 =− 15Therefore, for the f...

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MCV4U Unit 3: Curve Sketching 1a. Given: 2

𝑓(𝑥) = 𝑥 + 2𝑥 − 15 To find the x-intercepts, I set y=0 and solved for x: 2

𝑓(𝑥) = 𝑥 + 2𝑥 − 15 𝑓(𝑥) = (𝑥 + 5)(𝑥 − 3) 𝑥+ 5 = 0 𝑥 =− 5 𝑥− 3 = 0 𝑥= 3 To find the y-intercepts, I set x=0 and solved for y: 2

𝑓(0) = 0 + 2(0) − 15 𝑓(0) =− 15 𝑦 =− 15 2

Therefore, for the function 𝑓(𝑥) = 𝑥 + 2𝑥 − 15, the x-intercepts are (-5,0) (3,0) and the y-intercept is (0,-15).

1b. Given: 𝑦=

𝑥+10 𝑥−5

To find the x-intercepts, I set y=0 and solved for x: 0 =

𝑥+10 𝑥−5

𝑥 + 10 = 0 𝑥 =− 10 To find the y-intercepts, I set x=0 and solved for y: 𝑦=

𝑥+10 𝑥−5

𝑦=

0+10 0−5

𝑦 =− 2 Therefore, for the function 𝑦 =

𝑥+10 𝑥−5

,

the x-intercepts is(-10,0) and the y-intercept is (0,-2).

2a. Given: 2

𝑦=

𝑥 −3𝑥+2 𝑥−4

𝐷 = {𝑥 ϵ 𝑅/𝑥 ≠ 4}

2b. Given: 3

2

𝑦 = 4𝑥 + 7𝑥 − 13𝑥 + 100 𝐷 = {𝑥 ϵ 𝑅} 2c. Given: 𝑓(𝑥) = 2𝑥 − 10 2𝑥 − 10 ≥ 0 𝑥≥ 5 𝐷 = {𝑥 ϵ 𝑅/𝑥 ≥ 5} 3a. Given: 𝑓(𝑥) =

𝑓(𝑥) =

1 3

5 2

(𝑥 +𝑥 ) 1 6

8

10

𝑥 +2𝑥 +𝑥

𝑓(− 𝑥) = 𝑓(− 𝑥) =

1 6

8

10

(−𝑥) +2(−𝑥) +(−𝑥) 1 6

8

10

𝑥 +2𝑥 +𝑥 1

− 𝑓(𝑥) =−

6

8

10

𝑥 +2𝑥 +𝑥

This function is even because 𝑓(− 𝑥) = 𝑓(𝑥).

3b. Given: 5

3

𝑦=𝑥 +𝑥 + 7 5

3

𝑦 = (− 𝑥) + (− 𝑥) + 7 5

3

5

3

𝑦 =− 𝑥 − 𝑥 + 7 𝑦 =− 𝑥 − 𝑥 − 7 This function is neither even nor odd because 𝑓(𝑥) ≠ 𝑓(− 𝑥) ≠− 𝑓(𝑥). 3c. Given: 𝑓(𝑥) =

3 3

𝑥+𝑥

𝑓(− 𝑥) = 𝑓(− 𝑥) = 𝑓(− 𝑥) = − 𝑓(𝑥) =

3 3

(−𝑥)+(−𝑥) 3 3

−𝑥−𝑥 −3 3

𝑥+𝑥 −3

3

𝑥+𝑥

This function is odd because 𝑓(− 𝑥) =− 𝑓(𝑥).

4a. Given: 3

𝑦=

𝑥 +5𝑥−7 2

𝑥 −3𝑥−28 3

𝑦=

𝑥 +5𝑥−7 (𝑥−7)(𝑥+4)

𝑥 = 7, 𝑥 =− 4 Therefore, the vertical asymptotes are 𝑥 = 7 and 𝑥 =− 4. 4b. Given: 𝑓(𝑥) =

5

3

3

2

𝑥 +4𝑥 −4𝑥+6 𝑥 +2𝑥 −5𝑥−6

I tested x=1 as a factor first: 3

2

(− 1) + 2(− 1) − 5(− 1) − 6 = 0 Thus, x-1 is a factor: 3

2

𝑥 +2𝑥 −5𝑥−6 𝑥−1 2

𝑥 + 𝑥− 6 (𝑥 + 3)(𝑥 − 2) Thus, the vertical asymptotes are: 𝑥 =− 3, 𝑥 = 2, 𝑥 = 1

5a. Given: 3

𝑦=

2

𝑥 +2𝑥 −6 2

𝑥

There is no horizontal asymptote since the degree of the numerator is larger than that of the denominator. 5b. Given: 4

𝑦=

3

2

𝑥 +3𝑥 +𝑥 −7 5

3

𝑥 −3𝑥 +8

The horizontal asymptote is y=0 because the degree of the denominator is larger than that of the numerator. 5c. Given: 2

𝑓(𝑥) =

2𝑥 2

𝑥 +3𝑥−4

The horizontal asymptote is y=2, because the degree of the numerator and denominator is the same I divided the coefficients

2 1

for an answer of y=2.

6a. Given: 2

𝑓(𝑥) = 4𝑥 + 12𝑥 − 7 I first found the derivative of the given function: 𝑓(𝑥) = 8𝑥 + 12 To find the critical values, I set the derivative equal to zero and solved for x. I then substituted the x value into the original given function to find the y value: 8𝑥 + 12 = 0 8𝑥 =− 12 𝑥= 𝑓( 𝑓(

−3 2 −3 2

−12 8

=

) = 4(

−3 2 −3 2 ) 2

+ 12(

−3 ) 2

− 7

) =− 16

To find the intervals of increasing or decreasing, I made a table: Intervals

𝑥<

−3 2

8𝑥

8𝑥 − 12

Increasing or Decreasing

-

-

Decreasing

(𝑥 =

−5 2

)

𝑥=

−3 2

-

-

Decreasing

𝑥>

−3 2

+

+

Increasing

(𝑥 = 2) Therefore: The critical value is

−3 2

The function is decreasing in the interval 𝑥 < increasing in the interval 𝑥 > The minimum point is (

−3 2

−3 2

and

−3 2

, − 16)

6b. Given: 3

2

𝑓(𝑥) = 𝑥 − 9𝑥 + 24𝑥 − 10 I first found the derivative of the given function: 2

𝑓(𝑥) = 3𝑥 − 18𝑥 + 24 To find the critical values, I set the derivative equal to zero and solved for x. I then substituted the x values into the original given function to find the y values: 2

3𝑥 − 18𝑥 + 24 = 0

3(𝑥 − 2)(𝑥 − 4) = 0 𝑥 = 2, 𝑥 = 4 3

2

3

2

𝑓(2) = 2 − 9(2) + 24(2) − 10 𝑓(2) = 10 𝑓(4) = 2 − 9(4) + 24(4) − 10 𝑓(4) = 6 To find the intervals of increasing or decreasing, I made a table: Intervals

𝑥− 2 𝑥− 4

3(𝑥 − 2)(𝑥 − 4)

Increasing or Decreasing

𝑥< 2 (𝑥 = 1)

-

-

+

Increasing

2 < 𝑥< 4 + (𝑥 = 3)

-

-

Decreasing

+

+

+

Increasing

𝑥> 4 (𝑥 = 5)

Therefore: The critical values are 𝑥 = 2, 𝑎𝑛𝑑 𝑥 = 4

The function is increasing in the interval 𝑥 < 2 and 𝑥 > 4, and decreasing in the interval 2 < 𝑥 < 4. The maximum points is (2, 10) and the minimum point is (4, 6) 6c. Given: 2

𝑥

𝑦=

2

𝑥 +2𝑥−15

I first found the derivative of the given function: 2

𝑓(𝑥) =

2

(𝑥 −2𝑥−15)−(𝑥 )(2𝑥+2) 2

2

(𝑥 +2𝑥−15) 2

𝑓(𝑥) =

2𝑥 −30𝑥 2

2

(𝑥 +2𝑥−15)

To find the critical values, I set the derivative equal to zero and solved for x. I then substituted the x values into the original given function to find the y values: 2

2𝑥 −30𝑥 2

2

(𝑥 +2𝑥−15)

= 0

𝑥 = 0, 𝑥 = 15 𝑓(0) = 0 𝑓(15) =

15 16

2

𝑥 + 2𝑥 − 15 = (𝑥 + 5)(𝑥 − 3) 𝑥 =− 5, 𝑥 = 3

To find the intervals of increasing or decreasing, I made a table: 2

Intervals

𝑥+ 5 𝑥− 3

𝑥 + 2𝑥 − 15

Increasing or Decreasing

𝑥 15 (𝑥 = 20)

+

+

+

Increasing

Therefore: The critical values are 𝑥 = 0, 𝑎𝑛𝑑 𝑥 = 15

The function is increasing in the intervals 𝑥 15 , and decreasing in the interval − 5 < 𝑥 < 3. The maximum point is (0, 0) and the minimum point is (15,

15 16

)

7a. Given: 4

3

𝑓(𝑥) = 𝑥 − 2𝑥 + 𝑥 − 2 I first found the first and second derivative of the given function, and then set the second derivative to 0 and solved for x: 3

2

𝑓'(𝑥) = 4𝑥 − 6𝑥 + 1 2

𝑓''(𝑥) = 12𝑥 − 12𝑥 𝑓''(𝑥) = 0 2

12𝑥 − 12𝑥 = 0 𝑥(12𝑥 − 12) = 0 12𝑥 − 12 = 0 𝑥=

12 12

𝑥= 1 Thus, the inflection points are 𝑥 = 0, 𝑎𝑛𝑑 𝑥 = 1.

I then made a table to find concavity: Intervals

𝑥− 0

𝑥− 1

2 12𝑥 − 12𝑥 Concavity

𝑥< 0 (𝑥 =− 1)

-

-

+

Concave Up

0 < 𝑥< 1 + (𝑥 = 0. 5)

-

-

Concave Down

+

+

+

Concave Up

𝑥> 1 (𝑥 = 2)

Therefore, the points of inflection are 𝑥 = 0, 𝑎𝑛𝑑 𝑥 = 1, and the concavity is upwards during the intervals 𝑥 < 0 and 𝑥 = 1, but is downwards when 0 < 𝑥 < 1. 7b. Given: 2

𝑓(𝑥) =

𝑥 2

𝑥 −4

I first found the first and second derivative of the given function, and then set the second derivative to 0 and solved for x:

2

2

𝑓'(𝑥) =

(𝑥 −4)(2𝑥)−(𝑥 )(2𝑥)

𝑓'(𝑥) =

2𝑥(𝑥 −4)−2𝑥

2

2

(𝑥 −4) 2

𝑓'(𝑥) =

3

2

2

(𝑥 −4) (−8𝑥) 2

2

(𝑥 −4)

2

2

2

2

𝑥(𝑥 −4) −𝑥(𝑥 −4)

𝑓''(𝑥) =− 8 ·

2

2 2

((𝑥 −4) ) 2

2

2

2

𝑓''(𝑥) =

−8(1(𝑥 −4) −2(𝑥 −4)(𝑥 −4)(𝑥))

𝑓''(𝑥) =

−8((𝑥 −4) −2(𝑥 −4)(2𝑥+0)𝑥)

𝑓''(𝑥) =

−8((𝑥 −4) −4𝑥 (𝑥 −4))

𝑓''(𝑥) =

−8(−3𝑥 +4)

2

4

(𝑥 −4) 2

2

2

2

4

(𝑥 −4) 2

2

2

2

4

2

(𝑥 −4) 2

2

3

(𝑥 −4)

𝑓''(𝑥) = 0 2

− 8(− 3𝑥 − 4) = 0 2

4 =− 3𝑥 𝑥=

−4 3

Thus, because this is a negative root, there are no real roots, meaning there are no inflection points. So I instead had to find restrictions on the domain: 2

𝑥 − 4 = 0 (𝑥 − 2)(𝑥 + 2) = 0 𝑥 = 2, − 2

I then made a table to find concavity: 2

Intervals

−8(−3𝑥 +4) 2

Concavity

3

(𝑥 −4)

𝑥 2 (𝑥 = 3)

+

Concave Up

Therefore, there are no inflection points, and the function concaves up at the intervals 𝑥 2 but concaves down at the interval − 2 < 𝑥 < 2 8. Given: 4

2

𝑓(𝑥) = 𝑥 − 8𝑥 + 5 I first found the first derivative, set it equal to zero and solved for the x values: 3

𝑓'(𝑥) = 4𝑥 − 16𝑥 2

𝑓'(𝑥) = 4𝑥(𝑥 − 4) 𝑓'(𝑥) = 4𝑥(𝑥 − 2)(𝑥 + 2)

𝑓'(𝑥) = 0 4𝑥(𝑥 − 2)(𝑥 + 2) = 0 𝑥 = 2, 𝑥 =− 2, 𝑥 = 0 I then found the second derivative, substituted in my vales and solved: 2

𝑓''(𝑥) = 12𝑥 − 16 2

𝑓''(0) = 12(0) − 16 𝑓''(0) =− 16 2

𝑓''(2) = 12(2) − 16 𝑓''(2) =− 32 2

𝑓''(− 2) = 12(− 2) − 16 𝑓''(− 2) = 32 I then substituted the x values into the original function: 4

2

4

2

𝑓(0) = (0) − 8(0) + 5 𝑓(0) = 5 𝑓(2) = (2) − 8(2) + 5 𝑓(2) =− 11 4

2

𝑓(− 2) = (− 2) − 8(− 2) + 5 𝑓(− 2) =− 11 According to the second derivative test, since 𝑓''(0) < 0, 𝑥 = 0 is a maximum.

According to the second derivative test, since 𝑓''(2) > 0, 𝑓''(2) is a minimum. According to the second derivative test, since 𝑓''(− 2) > 0, 𝑓''(− 2) is a minimum. Therefore for the function , the minimums are: (2, − 11)(− 2, − 11), and the maximum is: (0, 5) 9a. If a function changes from a decreasing interval to an increasing interval, the conclusion can be made that there is a minimum between the intervals 9b. If 𝑙𝑖𝑚

+

𝑥→0

𝑓(𝑥) =− ∞ 𝑎𝑛𝑑 𝑙𝑖𝑚

−

𝑥→0

𝑓(𝑥) = ∞ then the

conclusion can be made that there is a vertical asymptote at 𝑥 = 0 10. Given: 2

𝑦 = 𝑥 − 4𝑥 + 21 occurs 𝑥 = 2 2

𝑦 = (2) − 4(2) + 21 𝑦 = 17 Therefore there is a minimum at (2, 17)

11. +

When finding one-sided limits, a 0 implies that the direction from which you approach zero is right, and that you are approaching the value of 0 plus a little bit. 12. A limit does not exist when the one sided limits are not equal, when the function does not approach a finite value, and when the function does not approach a particular value. 13. A function if said to be continuous if it meets 2 conditions: First, if 𝑙𝑖𝑚𝑥→𝑎𝑓(𝑥) exists and second if 𝑙𝑖𝑚𝑥→𝑎𝑓(𝑥) = 𝑓(𝑎)

14a. Given: 3

2

𝑦 = 𝑥 + 𝑥 − 20𝑥 To analyze the function I used the curve sketching procedure: 1. Domain The function is not rational or radical, meaning there are no restrictions on the domain. 𝐷 = (𝑥 ϵ 𝑅) 2. Intercepts 𝑦 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡: I set x=0 and solved for y: 3

2

𝑦 = (0) + (0) − 20(0) 𝑦 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = (0, 0) 𝑥 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠: I set y=0 and solved for x: 3

2

𝑥 + 𝑥 − 20𝑥 = 0 2

𝑥(𝑥 + 𝑥 − 20) = 0 𝑥(𝑥 − 4)(𝑥 + 5) = 0 𝑥 = 0, 𝑥 = 4, 𝑥 =− 5

𝑥 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 = 0, 4, − 5

3. Symmetry I tested both 𝑓(− 𝑥) 𝑎𝑛𝑑 − 𝑓(𝑥): 3

2

𝑓(− 𝑥) = (− 𝑥) + (− 𝑥 ) − 20(− 𝑥) 3

2

3

2

𝑓(− 𝑥) =− 𝑥 + 𝑥 + 20𝑥 − 𝑓(𝑥) =− 𝑥 − 𝑥 + 20𝑥 The function is neither even nor odd since 𝑓(𝑥) ≠ 𝑓(− 𝑥) ≠− 𝑓(𝑥). 4. Asymptotes No asymptotes since there are no restrictions on the domain, and the function is not rational. 5. Intervals of increase or decrease I found the first derivative, set it equal to zero and solved: 3

2

𝑓(𝑥) = 𝑥 + 𝑥 − 20𝑥 2

𝑓'(𝑥) = 3𝑥 + 2𝑥 − 20 2

3𝑥 + 2𝑥 − 20 = 0 This was unfactorable so I used the quadratic formula: 𝑎 = 3, 𝑏 = 3, 𝑐 =− 20 2

𝑥=

−2± (−2) −4(3)(−20) 2(6)

𝑥=

−1± 61 3

Thus the critical values are: 𝑥 = 2. 27008 𝑜𝑟 − 2. 93675 I then made a table: 2

Intervals

𝑥 − 2. 27

𝑥 + 2. 937

3𝑥 + 2𝑥 − 20

Increasing or Decreasing

𝑥 < 2. 27 (𝑥 = 1)

-

+

-

Decreasing

2. 27 < 𝑥 < 2. 937 + (𝑥 = 2. 5)

+

+

Increasing

+

+

+

Increasing

𝑥 > 2. 93675 (𝑥 = 4)

Thus, 𝑓(𝑥) is increasing when 2. 27 < 𝑥 < 2. 937 and 𝑥 > 2. 937, but decreasing when 𝑥 < 2. 27. 6. Maximum and minimum points 3

2

𝑓(− 2. 937) = (− 2. 937) + (− 2. 937) − 20(− 2. 937) 𝑓(− 2. 937) = 42. 03 3

2

𝑓(2. 27) = (2. 27) + (2. 27) − 20(2. 27) 𝑓(2. 27) =− 28. 55

Thus, the maximum is (− 2. 937, 42. 03) and the minimum is (2. 27, − 28. 55)

7. Intervals of concavity I found the second derivative, set it equal to zero and solved: 2

𝑓'(𝑥) = 3𝑥 + 2𝑥 − 20 𝑓''(𝑥) = 6𝑥 + 2 6𝑥 + 2 − 0 6𝑥 =− 2 𝑥=

−1 3

I set up a table for concavity: Intervals −1 3 −2 = 3

𝑥< (𝑥

−1 3 1 =3)

𝑥> (𝑥

6𝑥 + 2

Concavity

-

Concave Down

+

Concave Up

)

Therefore the functions concaves down when 𝑥 <

−1 3

and up when 𝑥 >

−1 3

.

8. Points of inflection The function changes from concave down to up, meaning there is an inflection point when 𝑥 = 𝑓( 𝑓( 𝑓( 𝑓(

−1 3 −1 3 −1 3 −1 3

−1 3 ) + 3 2 20 + 3 27 182 27

) = ( ) = ) =

(

−1 2 ) 3

− 20(

−1 3

−1 ) 3

) = 6. 74

Thus the inflection point is (

−1 3

, 6. 74)

:

9. Sketch the curve

14b. Given: 2

𝑦=

1+𝑥

2

1−𝑥

To analyze the function I used the curve sketching procedure: 1. Domain This function is a rational function, meaning there are restrictions on the domain and the denominator cannot be zero. 2

1 −𝑥 = 0 2

𝑥 = 1 𝑥 ≠ 1, − 1 𝐷 = (𝑥 ϵ 𝑅/𝑥 ≠ 1, − 1) 2. Intercepts 𝑦 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡: I set y=0 and solved

2

𝑦=

1+(0)

2

1−(0)

𝑦= 1 𝑦 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡 = (0, 1) 𝑥 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡: 𝑁𝑜 𝑥 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡𝑠 𝑏𝑒𝑐𝑎𝑢𝑠𝑒 𝑡ℎ𝑒 𝑑𝑒𝑛𝑜𝑚𝑖𝑛𝑎𝑡𝑜𝑟 𝑐𝑎𝑛𝑛𝑜𝑡 𝑏𝑒 𝑧𝑒𝑟𝑜. 3. Symmetry I tested 𝑓(− 𝑥) 𝑎𝑛𝑑 − 𝑓(𝑥) 2

𝑓(− 𝑥) =

1+(−𝑥)

𝑓(− 𝑥) =

1+𝑥

− 𝑓(𝑥) =

−1−𝑥

2

1−(−𝑥) 2 2

1−𝑥

2 2

−1+𝑥

The function is even since 𝑓(− 𝑥) = 𝑓(𝑥) 4. Asymptotes There are restriction on the domain therefore: 𝑉𝑒𝑟𝑡𝑖𝑐𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒𝑠: 𝑥 = 1, 𝑥 =− 1 This is a rational function, but the degrees are equal, therefore: 𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒: 𝑦=

1 −1

𝐻𝑜𝑟𝑖𝑧𝑜𝑛𝑡𝑎𝑙 𝑎𝑠𝑦𝑚𝑝𝑡𝑜𝑡𝑒𝑠: 𝑦 =− 1

5. Intervals of increase or decrease I found the first derivative, set equal to 0 and solved: 2

1+𝑥

𝑓(𝑥) =

2

1−𝑥

2

𝑓'(𝑥) = 𝑓'(𝑥) = 4𝑥 2 2

(1−𝑥 )

2

(1−𝑥 )(2𝑥)−(1+𝑥 )(−2𝑥) 2 2

(1−𝑥 ) 4𝑥 2 2

(1−𝑥 )

= 0

4𝑥 = 0 𝑥= 0 Thus, the critical point is 𝑥 = 0. I combined it with the domain restrictions and made a table:

4𝑥

Intervals

2 2

(1−𝑥 )

Increasing or Decreasing

𝑥 1 (𝑥 = 2)

+

Increasing

Therefore 𝑓(𝑥) is decreasing at 𝑥 1. 6. Maximum and minimum points Since the function changes from decreasing to increasing at 0 < 𝑥 < 1, there is a minimum: 𝑓(0) = 1 Thus there is a minimum at (0, 1) 7. Intervals of concavity I found the second derivative, set it equal to 0 and solved: 𝑓'(𝑥) =

4𝑥 2 2

(1−𝑥 )

2 2

2

𝑓''(𝑥) =

(1−𝑥 ) (4)−(4𝑥)(2(1−𝑥 )(−2𝑥))

𝑓''(𝑥) =

4(3𝑥 +1)

2 2 2

((1−𝑥 ) ) 2

2 3

(1−𝑥 )

𝑓''(𝑥) = 0 2

4(3𝑥 + 1) = 0 2

𝑥 = 𝑥=

−1 3 −1 3

I used the restrictions to set up my table since there is no real root:

2

Interval

4(3𝑥 + 1)

Concavity

𝑥 1 (𝑥 = 2)

+

Concave Up

The function concaves up at all intervals. 8. Points of inflection No points of inflection because there is no real root.

9. Sketch the curve...