Calculus 12 - rcc-jlo.weebly.com

Calculus 12Chapter 4 – Applications of Differentiation(Part 2)(Section B)

Dr. John LoRoyal Canadian College

2020-2021

6. Introduction to curve sketching

› An important application of differentiation in calculus is to allow us to sketch graphs of functions.

› To obtain a good (i.e., accurate) graph, the following pieces of information are very crucial:

1. Function form: ordered pairs, intercepts, domain, range, asymptotes, end behavior, symmetry

2. First derivative: critical values, intervals of trends, stationary points

3. Second derivative: inflection points, intervals of concavity

CHAPTER 4 - APPLICATIONS OF DIFFERENTIATION (PART 2)

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› We have learned many types of functions in PREC 10, 11 and 12 (i.e., linear, quadratic, reciprocal, radical, polynomial, rational, sinusoidal, exponential and logarithmic). You are expected to know clearly the characteristics of these functions and their graphs.

› Please review your notes for these classes if you have any troubles.


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7. First derivative test

› From chapter 3 we have learned that the derivative of a function 𝑓(𝑥) at 𝑥 gives the slope of the tangent line at this position.

› That means the value of 𝑓′(𝑥) tells us the orientation of the tangent line, and thus the local behavior of the function at 𝑥.


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𝑓′ 𝑥 = limℎ→0

𝑓 𝑥 + ℎ − 𝑓(𝑥)

ℎ

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› There are five possible outcomes:

1. 𝑓′ 𝑥 > 0: the function is increasing

2. 𝑓′ 𝑥 < 0: the function is decreasing

3. 𝑓′ 𝑥 = 0: the function levels off

4. 𝑓′ 𝑥 = ±∞: the function is going straight up or down

5. 𝑓′ 𝑥 = DNE: the function is not differentiable


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› Graphically:

› Based on this graph, we see that there exists a point through which the graph changes its tendency (i.e., from increasing to decreasing, or vice versa).


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› This point is called a critical point, and the associated value of 𝑥 is called a critical number (or critical value).

› It can be found by checking:

› The point 𝑥 at which 𝑓′ 𝑥 = 0 is also called a stationary point.


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𝑓′ 𝑥 = 0 𝑓′ 𝑥 = DNE

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› Example: Find the critical points on the graph of 𝑓 𝑥 =𝑥3 − 𝑥2 − 𝑥 + 2.

› Solution: The derivative of 𝑓(𝑥) is

› Setting 𝑓′ 𝑥 = 0 gives

› When 𝑥 = 1, 𝑓 1 = 1. When 𝑥 = −1/3, 𝑓 −1

3= 59/27.


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𝑓′ 𝑥 = 3𝑥2 − 2𝑥 − 1

3𝑥 + 1 𝑥 − 1 = 0

𝑥 = 1 or −1

3

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› We can describe the graph by stating the intervals in which the graph is increasing and the intervals in which the graph is decreasing.

1. The interval of increase:

– An interval where 𝑓′ 𝑥 > 0

2. The interval of decrease:

– An interval where 𝑓′ 𝑥 < 0


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› Example: Find the intervals of increase/decrease for 𝑓 𝑥 = 3𝑥4 + 4𝑥3 − 12𝑥2 + 2.

› Solution: First find the critical number(s):

› Now the number line can be split into several intervals. For each interval, we choose a point and evaluate the derivative.


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𝑓′ 𝑥 = 12𝑥3 + 12𝑥2 − 24𝑥 = 0

12𝑥 𝑥2 + 𝑥 − 2 = 0

12𝑥 𝑥 + 2 𝑥 − 1 = 0

𝑥 = 0, 1, −2

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› Look at the following number line:


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𝑓′ −3 = 12(−3)3+12(−3)2−24 −3 = −144 < 0

𝑓′ −1 = 12(−1)3+12(−1)2−24 −1 = 24 > 0

𝑓′ 0.5 = 12(0.5)3+12(0.5)2−24 0.5 = −10 < 0

𝑓′ 2 = 12(2)3+12(2)2−24 2 = 96 > 0

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› To summarize:


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Interval 𝑓′(𝑥) Trend

(−∞,−2) < 0 decreasing

(−2, 0) > 0 increasing

(0, 1) < 0 decreasing

(1,∞) > 0 increasing

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› To verify our answers, we plot the function 𝑓(𝑥):


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› From the previous example, we can summarize the following key points which constitute the first derivative test:

1. A critical point is a relative maximum if the derivative is positive to its immediate left and the derivative is negative to its immediate right.

2. A critical point is a relative minimum if the derivative is negative to its immediate left and the derivative is positive to its immediate right.

3. A critical point is neither a relative maximum nor a relative minimum if the derivative to its immediate left and to its immediate right are both positive or both negative.


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8. Second derivative test

› While the first derivative of a function indicates how it changes with 𝑥, the second derivative of the function tells how the value of 𝑓′(𝑥) changes with 𝑥. In other words, it tells the rate at which slope of the graph changes.


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𝑓′′ 𝑥 = lim∆𝑥→0

𝑓′ 𝑥 + ∆𝑥 − 𝑓′(𝑥)

∆𝑥

=𝑑

𝑑𝑥

𝑑𝑓

𝑑𝑥

First derivative of 𝑓(𝑥) → slope

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› Like the case of first derivative, second derivative calculations will yield one of the following outcomes:

1. Positive 𝑓′′(𝑥): slope is increasing; it implies upwardconcavity.

2. Negative 𝑓′′(𝑥): slope is decreasing; it implies downwardconcavity.

3. Zero 𝑓′′(𝑥): slope is neither increasing nor decreasing; it implies the graph is changing its concavity.

4. Undefined 𝑓′′(𝑥): either 𝑓′(𝑥) is undefined or the slope jumps.


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› Graphically:

› The point where the graph changes its concavity from up to down, or vice versa, is called an inflection point.

› At this point:


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𝑓′′ 𝑥 = 0 𝑓′′ 𝑥 = DNE

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› Example: Find the inflection points on the graph of 𝑓 𝑥 =𝑥3 − 𝑥2 − 𝑥 + 2.

› Solution: We first find the second derivative:

› Setting 𝑓′′ 𝑥 = 0 yields

› When 𝑥 =1

3, 𝑦 =

43

27.


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𝑓′ 𝑥 = 3𝑥2 − 2𝑥 − 1

𝑓′′ 𝑥 = 6𝑥 − 2

𝑥 =1

3

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› Using the inflection points, we are able to determine the intervals of concavity, in which the type of concavity of a function does not change.

1. The interval of “concave up”

– The second derivative is positive for all 𝑥 values.

2. The interval of “concave down”

– The second derivative is negative for all 𝑥 values.


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› Example: Find the intervals of concavity for

› Solution: The second derivative of the function is

› The inflection point is at 𝑥 = 1/3.

› We therefore consider two intervals:

› For 𝑥 < 1/3:

› For 𝑥 > 1/3:


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𝑓 𝑥 = 𝑥3 − 𝑥2 − 𝑥 + 2

𝑓′′ 𝑥 = 6𝑥 − 2

𝑓′′ 0 = −2 < 0

𝑓′′ 1 = 6 − 2 = 4 > 0

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› Therefore, we can put this information together on the number line:

› Summary:


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Intervals Concavity Local Extrema

(−∞,1

3)

downward maximum

(1

3,∞)

upward minimum

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› We can summarize the relationship between the local extrema and the concavity of a function as the second derivative test:

1. A critical point is a relative maximum if the second derivative is negative there.

2. A critical point is a relative minimum if the second derivative is positive there.

3. If the second derivative is zero or undefined, the critical point could be a relative maximum, a relative minimum, or neither. Additional tests have to be done to verify its identity.


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9. Practice on curve sketching

› The suggested procedures of sketching the graph of a function are as follows:

1. List out the general properties of the function; i.e., intercepts, symmetry, asymptotes, domain, and range.

2. Perform the 1st derivative test to find critical points. Determine the trend of the graph in different intervals.

3. Perform the 2nd derivative test to find inflection points and determine the concavity in different intervals. Use the results to confirm relative extrema found in step 2.


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› Example: Sketch 𝑓 𝑥 = 𝑥5 − 4𝑥3.

› Solution: We first look at the general properties of 𝑓(𝑥).

› Domain: (−∞,∞)

› Range: −∞,∞

› The 𝑦-intercept: (0, 0)

› The 𝑥-intercepts: We need to solve the polynomial equation:

› Hence, −2, 0 , (0, 0), (2, 0) [(0, 0) is a triple root]


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𝑥5 − 4𝑥3 = 0

𝑥3 𝑥2 − 4 = 0

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› The end behavior: From QIII to QI

› Asymptotes: None

› Now we can perform the 1st derivative test:

› The critical values are therefore 0 and ± 12/5.

› Then check the trend of 𝑓(𝑥) in different intervals:


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𝑓′ 𝑥 = 5𝑥4 − 12𝑥2 = 0

𝑥2 5𝑥2 − 12 = 0

𝑥 < −1.549 −1.549 < 𝑥 < 0 0 < 𝑥 < 1.549 𝑥 > 1.549

𝑓′ −2 = 32 𝑓′ −1 = −7 𝑓′ 1 = −7 𝑓′ 2 = 32

Increasing Decreasing Decreasing Increasing

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› We know that then 𝑥 = − 12/5 is a maximum and 𝑥 =

12/5 is a minimum. But we don’t know what is 𝑥 = 0.

› To answer this question, we perform the 2nd derivative test.

› Since 𝑓′′ 0 = 0, 𝑥 = 0 is a stationary, inflection point.


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𝑓′′ 𝑥 = 20𝑥3 − 24𝑥 = 0

4𝑥 5𝑥2 − 6 = 0

𝑥 = 0,± 6/5

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› The other two points will be typical inflection points where the function changes its concavity.

› Using this information, we would be able to sketch the graph of 𝑓(𝑥).


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𝑥 < −1.095 −1.095 < 𝑥 < 0 0 < 𝑥 < 1.095 𝑥 > 1.095

𝑓′′ −2 = −112 𝑓′′ −1 = 4 𝑓′′ 1 = −4 𝑓′′ 2 = 112

Concave down Concave up Concave down Concave up

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Maximum

Minimum

Inflection points

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› Example: Sketch

› Solution: We first find out the domain of the function. Note that 𝑥2 − 4 = 0 gives 𝑥 = ±2. Hence

› This also tells us that there are two vertical asymptotes:

› This is a rational function with deg 𝑝 𝑥 = deg(𝑞 𝑥 ).


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𝑦 =2 𝑥2 − 9

𝑥2 − 4

𝐷: −∞,−2 ∩ −2, 2 ∩ 2,∞

V. A. : 𝑥 = 2, 𝑥 = −2

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› Therefore,

› Now we can check for intercepts. Setting 𝑥 = 0 yields

› Setting 𝑦 = 0 gives


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H. A.= lim𝑥→∞

2 𝑥2 − 9

𝑥2 − 4= lim

𝑥→∞

2 1 −9𝑥2

1 −4𝑥2

= 2

𝑦 =2 0 − 9

0 − 4= 4.5

𝑥2 − 9 = 0

𝑥 = ±3

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› To find the critical values, we differentiate 𝑦:

› When 𝑦′ = 0, 𝑥 = 0. Hence there is only one stationary point. Meanwhile, there are two vertical asymptotes which are critical points too.


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𝑦′ =2 𝑥2 − 4 2𝑥 − 2 𝑥2 − 9 2𝑥

𝑥2 − 4 2=

20𝑥

𝑥2 − 4 2

(−∞,−2) (−2, 0) (0, 2) (2,∞)

𝑓′ −3 = −2.4 𝑓′ −1 = −2.2 𝑓′ 1 = 2.2 𝑓′ 3 = 2.4

Decreasing Decreasing Increasing Increasing

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› Take the second derivative of 𝑦:

› There are two inflection points, 𝑥 = ±2, which correspond to the vertical asymptotes.


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𝑦′′ =𝑥2 − 4 2 20 − 20𝑥 (2) 𝑥2 − 4 (2𝑥)

𝑥2 − 4 4

=20 𝑥2 − 4 𝑥2 − 4 − 4𝑥2

𝑥2 − 4 4

=−20 3𝑥2 + 4

𝑥2 − 4 3

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› To check the concavity:

› Based on these, we can conclude that 𝑥 = 0 should be a relative minimum. There exists no relative maximum.

› Moreover, we can deduce that the range of 𝑦 should be as follows:


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(−∞,−2) (−2, 2) (2,∞)

𝑓′′ −3 = −4.96 𝑓′′ 0 = 1.25 𝑓′′ 3 = −4.96

Concave down Concave up Concave down

𝑅: −∞, 2 ∩ 4.5,∞

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› Finally, we can plot the graph.


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› Example: Sketch the graph of 𝑦 =𝑥2+1

𝑥.

› Solution: Note that the function is undefined when 𝑥 = 0. That means

› There exists one vertical asymptote:

› This function does not have 𝑦-intercept since it is undefined when 𝑥 = 0.

› Similarly, it has no 𝑥-intercept since 𝑥2 + 1 = 0 has no solutions.


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𝐷: −∞, 0 ∩ 0,∞

𝑥 = 0

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› When 𝑥 → ±∞,

› Hence, it has an oblique asymptote whose equation is given by

› The graph is based on 𝑦 = 1/𝑥 with an oblique asymptote of 𝑦 = 𝑥.


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lim𝑥→±∞

𝑥2 + 1

𝑥= lim

𝑥→±∞𝑥 +

1

𝑥= ±∞

𝑦 =𝑥2 + 1

𝑥= 𝑥 +

1

𝑥

Oblique asymptote: 𝑦 = 𝑥

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› The first derivative of 𝑦:

› The critical values are 𝑥 = 0,±1.

› When 𝑥 = ±1, 𝑦 = 2. Hence, the critical points are (±1, 2).

› Now we can check the trends:


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𝑦′ =𝑥 2𝑥 − 𝑥2 + 1 (1)

𝑥2=𝑥2 − 1

𝑥2

−∞,−1 −1, 0 (0, 1) (1,∞)

𝑓′ −2 > 0 𝑓′ −0.5 < 0 𝑓′ 0.5 < 0 𝑓′ 2 > 0

Increasing Decreasing Decreasing Increasing

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› The second derivative of 𝑦:

› Since 𝑦′′ ≠ 0 for all 𝑥, yet 𝑥 = 0 makes 𝑓′′ undefined, this function has no inflection point.

› Now check for concavity:


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𝑦′′ =𝑥2 2𝑥 − 𝑥2 − 1 (2𝑥)

𝑥4=2𝑥

𝑥4=

2

𝑥3

(−∞, 0) (0,∞)

𝑓′′ −1 < 0 𝑓′′ 1 > 0

Concave down Concave up

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› Using the information we can plot the graph.


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› Example: Sketch 𝑦 = 3𝑥5/3 − 15𝑥2/3.

› Solution: This function is defined everywhere; therefore its domain is −∞,∞ . It implies there is no vertical asymptote.

› The 𝑦-intercept is:

› To find the 𝑥-intercepts it needs a little bit trick:

› When 𝑦 = 0, 𝑥 = 0, 5. CHAPTER 4 - APPLICATIONS OF DIFFERENTIATION

(PART 2)40

𝑦 = 3(0)5/3−15 023 = 0

𝑦 = 3𝑥5/3 − 15𝑥2/3

= 3𝑥2/3 𝑥 − 5

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› Since 𝑥5/3 increases faster than 𝑥2/3, when 𝑥 → ∞, 𝑦 →∞.

› Now find the first derivative of 𝑦:

› Setting 𝑦′ = 0 yields

› It gives a stationary point of 2. Note that 𝑥 = 0 makes 𝑦′undefined. It is therefore a critical point.


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𝑦′ = 5𝑥2/3 − 10𝑥−1/3

5𝑥−1/3 𝑥 − 2 = 0

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› We can check the trends:

› Then we look for the concavity. First we deduce the second derivative:


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(−∞, 0) (0, 2) (2,∞)

𝑓′ −1 > 0 𝑓′ 1 < 0 𝑓′ 3 > 0

Increasing Decreasing Increasing

𝑦′′ =10

3𝑥−1/3 +

10

3𝑥−4/3

=10

3𝑥−4/3 𝑥 + 1

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› Setting 𝑦′′ = 0:

› We can determine the inflection points: 0,−1.

› Using these we can work out the concavity:

› Compared with the trend table, we know that 𝑥 = 2 is a local minimum, yet 𝑥 = 0 is a cusp.


43

10

3𝑥−4/3 𝑥 + 1 = 0

(−∞,−1) (−1, 0) (0,∞)

𝑓′′ −2 < 0 𝑓′′ −0.5 > 0 𝑓′′ 1 > 0

Concave down Concave up Concave up

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› Finally we can sketch the graph:


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› Practice: Sketch the graph for the following functions.


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𝑦 =𝑥2 − 2𝑥 + 4

𝑥 − 2

𝑦 = 2𝑥5/3 − 5𝑥4/3

𝑦 = 𝑥4 − 12𝑥3 + 48𝑥2 − 64𝑥

𝑦 =𝑥

𝑥2 + 1

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𝑦 =𝑥2 − 2𝑥 + 4

𝑥 − 2

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𝑦 = 2𝑥5/3 − 5𝑥4/3

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𝑦 = 𝑥4 − 12𝑥3 + 48𝑥2 − 64𝑥

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𝑦 =𝑥

𝑥2 + 1

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