Calculus Chapter 3

Derivatives in Graphing and ApplicationsCalculus-I Chapter 4

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4.1 Analysis of Functions- I (PG:232)Increase, Decrease and Concavity

Remember-To find Increasing/Decreasing functions, always Read from Left to Right-

4.2 Analysis of Functions-II (pg:244)Relative Extrema;

Graphing Polynomial

Slope of a Tangent line dy/dx

Rate of change and slope

Secant line = Average Rate of Change = Slope

Atangent lineis a straight line that touches a function at only one point. (See above.) The tangent line represents theinstantaneous rate of changeof the function at that one point. The slope of thetangent lineat a point on the function is equal to thederivativeof the function at the same point (See below.)

In this first animation we see the secant line become the tangent line i.e we go from the Average Rate of Change to the Instantaneous Rate of Change by letting the interval over which the Average Rate of Changeis measured go to zero.

Secant line Tangent Line

Tangent Line = Instantaneous Rate of Change = Derivative

Finding SLOPE of a lineHave you ever heard of Mr. SLOPE GUY?

Mnemonics- Example: To find slopesFor horizontal lines:Slope=0Z for zero

For vertical lines:Slope is undefinedN for No slope

Memorizing the formulas: Use- Pascals Triangle

Increasing and Decreasing FunctionsDefinition: 4.1.1 (pg: 233)

Increasing and Decreasing Functions Theorem 4.1.2 (pg:233)Let f be a function that is continuous on a closed interval [a, b] and differentiable on the open interval (a, b).If f(x) > 0 for every x in (a, b), then f is increasing on (a,b) If f(x) < 0 for every x in (a, b), then f is decreasing on (a,b)If f(x) = 0 for every x in (a, b), then f is constant on (a,b)

Discontinuous function

Example Use the graph below to make a conjecture about the intervals on which f is increasing or decreasing.

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Use increasing/decreasing test to verify your conjecture.

-2 01 - - - - -+ + +- - - -+ + +Increasing:

Decreasing:

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Continuous functionFind the intervals on which the function f(x) is increasing and where it is decreasing.

Find the intervals on which the function f(x) is increasing and where it is decreasing.

4.2 CRITICAL POINTS (pg:245)Critical Values: The x-values at which the f (x)=0 or f (x) fails to exist.Note: 1. The x-values at which the f (x)=0 are called Stationary Points.Note: The critical values are the points where the graph will switch from increasing to decreasing or vice versa.Find the critical points for the following functions:Examples: 1.f(x) = 3x2 6x + 32. f(x) = x3/2 3x + 7x = 1x = 4

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Finding points where horizontal tangents to a curve occur

4.2 Exp: 3 (pg:245) Find all critical points of

Qs: 7 -14 Ex: 4.2 (pg: 252)Locate the critical points and identify which critical points are stationary points.Qs: 8

Qs: 9

Key TopicsRelative maximum: the highest value for f(x) at that particular peak in the graph

Relative minimum: the lowest value for f(x) at that particular valley in the graph

Relative maximumRelative maximumRelative minimumRelative minimumRelative minimum

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Key TopicsHow to determine whether it is a relative maximum or a relative minimum at a focal point:Step 1: Find the points of the graph to determine the intervals on which f(x) is increasing or decreasingStep 2: Choose an x-value in each interval to determine whether the function is increasing or decreasing within that intervalStep 3: If f(x) switches from increasing to decreasing at a critical point, there is a relative maximum at that critical point

If f(x) switches from decreasing to increasing at a critical point, there is a relative minimum at that critical point.

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Key TopicsIt might help to make a number line displaying your findings

- - - | +++++++ | - - - - | +++++++ | - - - - - - - | +++

- to + means minimum+ to - means maximum

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First derivative test to check for maximum and minimum pointsLet f be a twice differentiable function.Find f(x), and put it equal to zero. Find the critical points. Take test points, to check if f(x) > 0 or < 0, for increasing and decreasing function..If f (x) changes sign from + to -, then x = c has a maximum value, and if f changes sign from - to +, then x = c has a minimum value. If f(x) remains positive, then f is an increasing function, and if it remains negative, it is a deceasing function.

Relative Maxima and Minima

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Exp: 5 (pg: 248)IntervalTest valuesSign of f(x)Conclusion

Find the Relative Extrema of .Hint: First find critical points using f(x)=0.Then find intervals where the function is inc/dec.

Lets make a table:

4.1 Concavity- Up and Down (pg:235)CONCAVE UPCONCAVE DOWNA function f is concave up on an open interval if its tangent lines have increasing slopes on that interval. In this case, the graph of f lies above its tangent lines.A function f is concave down on an open interval if its tangent lines have decreasing slopes on that interval. In this case, the graph of f lies below its tangent lines.

concave up: f is increasing.tangent lines arebelow the graph.

concave down: f is decreasing.tangent lines areabove the graph.

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Concavity

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Mathematical Rhymes?

To determine the concavity of a function using second derivativeCONCAVITY

Concave upward

Concave downward

Think of some examples

Finding Concavity Theorem 4.1.4 (pg:235)Let f be a function which is twice differentiable on an open interval (a, b).If f(x) > 0 for every x in (a, b), then f is concave up on (a,b) If f(x) < 0 for every x in (a, b), then f is concave down on (a,b)

4.1 Inflection Points (pg:236)

If f is continuous on an open interval containing a value x = d, and if f changes the direction of its concavity at the point (d, f(d)), then f has an inflection point at x = d.

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Inflection PointTo find inflection points, find any point, c, in the domain where

changes sign from the left to the right of c,is undefined.

IfThen (c, f(c)) is an inflection point of f.

Find point of inflection.

ConcavityTest for concavity Find the second derivative. Put f (x) = 0 (to find points of inflection). Take test points.

If for every value of x in I, then f is concave up on I.

If for every value of x in I, then f is concave down on I.

Second derivative test to check for Concavity, and point of inflectionLet f be a twice differentiable function.Find f(x), and put it equal to zero. Find the points of inflection x= a. Take test points, to check if f(a) > 0 for concave up or f(a) < 0, for concave down.

Note: Find f(x); is undefined, gives points of inflection.

Example Given , find the intervals on which f is increasing/decreasing and concave up/down. Locate all points of inflection.

Solution

at x = 0 and at x = 202

Increasing: (-, 0) U (2, )Decreasing: (0, 2)

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Sign analysis of

1

This will tell us where it is concave up and down:

f (x) = 0

at x = 1

Concave Up:

Concave Down:

Questions: Find the interval where f is increasing/decreasing/concave up/concave down/ point of inflections, if any.Q:1

Q: 2

Q: 3

Q:4

4.2 Second Derivative Test: (pg:247)

4.2 Second Derivative Test: (pg:247)Suppose f is twice differentiable at the point x = c.(i) If f (x) = o and f (x) > 0, then f has a relative minimum at x = c.(ii) If f (x) = o and f (x) < 0, then f has a relative maximum at x = c.(iii) If f (x) = o and f (x) = 0, then the test is inconclusive; that is, f may have a relative maximum, a relative minimum, or neither at x = c.

Exp: Find the relative extrema using second derivative test: f(x)< 0 ---Max and f(x)> 0 ---Min IntervalTest valuesSign of f(x)Conclusion

Test for ConcavityIf f(x) > 0 for all x in I, then f is concave up.If f(x) < 0 for all x in I, then f is concave down.2nd Derivative Test (Use C.pts. of 1st Derivative)If f(c) > 0, then f(c) is a relative min.If f(c) < 0, then f(c) is a relative max.If f(c) = 0, then the test fails. No min. or max.

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Ex. 2Determine where f(x) = x4 4x3 is increasing, decreasing, has maxs or mins, is concave up or down, has inflection points.

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Ques: 54 Ex:4.2 (pg: 253)Draw the graph of the polynomial, and label the coordinates of the intercepts, stationary points, and inflection points.Check your work with a graphing utility (Desmos).

For Graphing the function:

Collect all the information :1. Critical points: f '(x) = 0.2. Increasing / Decreasing intervals;3. Points of inflection: f"(x) = 04. Concavity: up/down5. End behaviour: Limits as x tends to plus / minus infinity ( )6. x- intercepts (put y = 0) and y intercepts (put x = 0).

Ex. 1Determine where f(x) = 6(x2 + 3)-1 is increasing, decreasing, has maxs or mins, is concave up or down, has inflection points.f(x) = -6(x2 + 3)-2(2x)

C.point x= 0

0+

1st der.test(0,2)max.

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C.N.s -1, 1

-1 1

concave up

down

up

Inflection points(-1, )(1, )

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2nd Derivative TestPlug C.N.s of 1st der. into 2ndderivative.C.N. from 1st der. was 0.

0 is a maximumRemember, a neg. in the 2nd der. means concave down.

Therefore, the point is a maximum.

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(0,2) max(-1, 3/2) Inf. pt.(1, 3/2) Inf. pt.There are no x-intercepts

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4.3 Graphs of Rational functions: (pg:254)

Properties of Graphs:1. Symmetries (replace x by -x and y by -y)2. x- intercepts (put y = 0) and y- intercepts (put x = 0)3. Relative extrema (maxima and minima)4. Concavity (up or down)5. Intervals of increase and decrease ( )6. Inflection points (Put and take test points)7. Asymptotes (Horizontal: Take limit as ) (Vertical: Put denominator = 0, after factoring and cancellation)8. End behavior as

Example: pg:255Sketch a graph of (Then check and confirm on Desmos)

A Classic ProblemYou have 40 feet of fence to enclose a rectangular garden along the side of a barn. What is the maximum area that you can enclose?Example 1

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What dimensions for a one liter cylindrical can will use the least amount of material? (Least amount of surface area)We can minimize the material by minimizing the surface area.

area ofends

lateralareaWe need another equation that relates r and h:

MotorOilExample 2Now plug in for h

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Area and Perimeter Example

Find the dimensions of a rectangular area of 225 square meters that has the least perimeter.Drawing a figure might help.

lwA = l w = 225P = 2 l + 2 wFrom the area equation solve for l and substitute

P = 2 l + 2 w =

There is a critical value at 15 and -15So w = 15 meters, and l = 15 meters(square area)

Perimeter = 60 m

Maximizing Revenue and ProfitA company manufactures and sells x television sets per month. The monthly cost and price-demand equations are:

R (x) = xp = C (x) = 60,000 + 60x and p = 200 x/50 a. Find the maximum revenue. Maximum at R(x) = 0

R (x) =

x =5,000 R(5000) = $500,000

Application #1 continued

P (x) = b. Find the maximum profit, the production level that will realize the maximum profit, and the price the company should charge for each television set. P (x) has a critical value at x =

P (3500) =p (3500) =C (x) = 60,000 + 60x and p = 200 x/50 P (x) =

3500.$185,000$130 per TV.

R (x) =