Loading in 2 Seconds...

Calculus Date: 12/18/13 Obj : SWBAT apply first derivative test

Loading in 2 Seconds...

- By
**kane** - Follow User

- 120 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about 'Calculus Date: 12/18/13 Obj : SWBAT apply first derivative test' - kane

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

Obj: SWBAT apply first derivative

test

http://youtu.be/PBKnttVMbV4 first derivative test inc.dec.

Today –ICover the second derivative test (easier than 1st)

https://www.math.ucdavis.edu/~kouba/CalcOneDIRECTORY/graphingdirectory/

n class: Start WS3-3A; complete for homework

Tomorrow Friday: With Christian complete any

remaining worksheets, make sure

You understand material from the

test.,etc.

- Announcements:
- Break Packet online on Friday
- Merry Christmas if I don’t see you

"Do not judge me by my successes,

judge me by how many times

I fell down and got back up again.“

Nelson Mandela

The first derivative test tells whether the

function, f(x), is increasing, decreasing or constant.

The second derivative test tells whether the

derivative, f’(x),is increasing, decreasing or constant,

or whether a function is concave up or concave down

and whether a point is a local max or min.

Concave upward

y

y

x

x

- Look at these two graphs. Each is concave upward, but one is decreasing and the other is increasing. We need to be able to determine concavity from the function and not just from the graph.

Concave downward

y

y

x

x

- Look at these two graphs. Each is concave downward, but one is decreasing and the other is increasing. We need to be able to determine concavity from the function and not just from the graph.

Concavity

Let f be a differentiable function on (a, b).

1.f is concave upward on (a, b) if f' is increasing on aa(a, b). That is f ''(x)0 for each value of x in (a, b).

2.f is concave downward on (a, b) if f' is decreasing on (a, b). That is f ''(x)0 for each value of x in (a, b).

concave upward

concave downward

Inflection Point

A point on the graph of f at which fis continuousandconcavity changes is called an inflection point.

To search for inflection points, find any point, c in the domain where f ''(x)0 or f ''(x)is undefined. Some call these points “hypercritical points”.

If f ''changes sign from the left to the right of c, then (c,f (c))is an inflection point of f.

The Second Derivative Test

Determine the sign of the second derivative of f to the left and right of the hypercritical point to determine if the function is concave up or concave down or neither.

left

right

conclusion

f (c) is an inflection point

up down

down up

f (c) is an inflection point

No change

Concavity remains the same

Example: Inflection Points

Find all inflection points of

Relative max.

f (0) = 1

The First and Second Derivative TestGraph and find all the relative extrema of

= 0 x = 2

- 2 +

+ 0 - 0 +

0 4

f(x) is concave down for all x in (-∞, 2) because f” < 0. There is an

Inflection point at (2, -15) f(x) = 0 and f(x) becomes concave upward

The First and Second Derivative Tests

f(x) is concave down for all x in (-∞, 2)

because f” < 0. There is an inflection point

at (2, -15) and f(x) =0 and where f(x)

becomes concave upward

What else can we say about

the rate of change?

There is a local maximum at (0,1)because >0 for all x in (-∞, 0) and < 0 for all x in (0,2).

There is a local minimum at (4,-31)because <0 for all x in (0,4)and > 0 for all x in (4, ∞).

Also the Second Derivative Test

can be used to find local extrema

Let c be a critical point of f(x). If f”(c) exists, then

- If f ''(c) > 0 then f(c) is a local minimum.
- If f ''(c) < 0 then f(c) is a local maximum.
- If f ''(c) = 0 inconclusive, use the first derivative test to classify the extrema.

Let c be a critical point of f(x). If f”(c) exists, then

- If f ''(c) > 0 then f(c) is a local minimum.
- If f ''(c) < 0 then f(c) is a local maximum.
- If f ''(c) = 0 inconclusive, use the first derivative test to classify the extrema.

Using the function below, find any relative extrema

f(0)= -12 f(4) = 12

Graph

- Take the derivative f’(x)
- Find the critical points f’(x) = 0; f’(x) = DNE
- Make sign chart
- Label critical points – put 0 or DNE on graph
- Evaluate the derivative at points on either side of extrema to determine the sign. Use Yvars for faster calculations. Mark the chart as + or – in these areas
- Evaluate f(x) at the critical points to determine actual values of extrema.
- Add arrows to show increasing or decreasing regions in f(x)
- Write out all extrema and use the value of the derivative and the appropriate interval to justify your answer.

The Point of Diminishing Returns

If the function represents the total sales of a particular object, t months after being introduced, find the point of diminishing returns.

S concave up on

S concave down on

The point of diminishing returns is at 20 months (the rate at which units are sold starts to drop).

Singular points:

Evaluate the derivative at points on either

side of extrema to determine the sign.

Use Yvars for

faster calculations

ND – derivative not defined

Relative max.

Relative min.

+ ND + 0 - ND - 0 + ND +

-1 0 1

Possibleextremaat .

Set

Example:

Graph

We can use a chart to organize our thoughts.

First derivative test:

y y=4 y=0

positive

negative

positive

Possible extreme at .

Set

maximum at

minimum at

Example:

Graph

First derivative test:

y y=4 y=0

There is a local maximum at (0,4) because for all x in

and for all x in (0,2) .

There is a local minimum at (2,0) because for all x in

(0,2) and for all x in .

Example:

Graph

NOTE: On the AP Exam, it is not sufficient to simply draw the chart and write the answer. You must give a written explanation!

First derivative test:

There is a local maximum at (0,4) because for all x in

and for all x in (0,2) .

There is a local minimum at (2,0) because for all x in

(0,2) and for all x in .

Example:

Graph

First derivative test:

Applications of the Derivative

- Maxima and Minima
- Applications of Maxima and Minima
- The Second Derivative - Analyzing Graphs

Absolute Extrema

A function f has an absolute (global) maximum atx = c if f (x) f (c)for allx in the domain D of f.

The number f (c) is called the absolute maximumvalue of f in D

Absolute Maximum

Absolute Extrema

A function f has an absolute (global) minimum atx = c if f (c) f (x)for allx in the domain D of f.

The number f (c) is called the absolute minimumvalue of f in D

Absolute Minimum

Relative Extrema

A function f has a relative (local) maximum at xc if there exists an open interval (r, s) containing c suchthat f (x) f (c) for all r x s.

Relative Maxima

Relative Extrema

A function f has a relative (local) minimum at xc if there exists an open interval (r, s) containing c suchthat f (c) f (x) for all r x s.

Relative Minima

Generic Example

The corresponding values of x are called Critical Points of f

Critical Points of f

A critical number of a function f is a number cin the domain off such that

(stationary point)

(singular point)

Candidates for Relative Extrema

- Stationary points: any x such that xis in the domain of f and f'(x) 0.
- Singular points: any x such that xis in the domain of f and f'(x) undefined
- Remark:notice that not every critical number correspond to a local maximum or local minimum. We use “local extrema” to refer to either a max or a min.

Fermat’s Theorem

If a function f has a local maximum or minimum at c, then c is a critical number of f

Notice that the theorem does not say that at every critical number the function has a local maximum or local minimum

Extreme Value Theorem

If a function f is continuous on a closed interval [a,b], then f attains an absolute maximum and absolute minimum on [a, b]. Each extremum occurs at a critical number or at an endpoint.

a b

a b

a b

Attains max. and min.

Attains min. but no max.

No min. and no max.

Open Interval

Not continuous

Absolute Min.

Absolute Max.

ExampleFind the absolute extrema of

Critical values of f inside the interval (-1/2,3) are x = 0, 2

Evaluate

Example

Find the absolute extrema of

Critical values of f inside the interval (-1/2,3) are x = 0, 2

Absolute Max.

Absolute Min.

Example

Find the absolute extrema of

Critical values of f inside the interval (-1/2,1) is x = 0 only

Absolute Max.

Evaluate

Absolute Min.

Example

Find the absolute extrema of

Critical values of f inside the interval (-1/2,1) is x = 0 only

Absolute Max.

Absolute Min.

Start Herea. Reviewing Rolle’s and MVTb. Remember nDeriv and Yvarsc. Increasing, Decreasing,Constantd. First Derivative Test

Finding absolute extrema on [a,b]

0. Verify function is continuous on the interval.

Determine the function’s domain.

- Find all critical numbers for f (x) in (a,b).
- Evaluate f (x) for all critical numbers in (a,b).
- Evaluate f (x) for the endpoints a and b of the interval [a,b].
- The largest value found in steps 2 and 3 is the absolute maximum for f on the interval [a , b], and the smallest value found is the absolute minimum for f on [a,b].

Rolle’s Theorem

- Given f(x) on closed interval [a, b]
- Differentiable on open interval (a, b)
- If f(a) = f(b) … then
- There exists at least one numbera < c < b such that f ’(c) = 0

f(a) = f(b)

b

a

aka the ‘crooked’ Rolle’s Theorem

f(b)

a

c

b

f(a)

- We can “tilt” the picture of Rolle’s Theorem
- Stipulating that f(a) ≠ f(b)

If f is continuous on [a, b] and differentiable on (a, b)

There is at least one number c on (a, b) at which

Conclusion:

Slope of Secant Line

Equals

Slope of Tangent Line

How is Rolle’s

Connected to MVT?

aka the ‘crooked’ Rolle’s Theorem

f(b)

a

c

b

f(a)

- We can “tilt” the picture of Rolle’s Theorem
- Stipulating that f(a) ≠ f(b)

If f is continuous on [a, b] and differentiable on (a, b)

There is at least one number c on (a, b) at which

Conclusion:

The average rate of change

equals the instantaneous rate of change

evaluated at a point

Finding c

- Given a function f(x) = 2x3 – x2
- Find all points on the interval [0, 2] where
- Rolle’s?
- Strategy
- Find slope of line from f(0) to f(2)
- Find f ‘(x)
- Set f ‘(x) equal to slope … solve for x

If , how many numbers on [-2, 3] satisfy

the conclusion of the Mean Value Theorem.

A. 0 B. 1 C. 2 D. 3 E. 4

CALCULATOR REQUIRED

f(3) = 39 f(-2) = 64

For how many value(s) of c is f ‘ (c ) = -5?

X

X

X

Given the graph of f(x) below, use the graph of f to estimate the

numbers on [0, 3.5] which satisfy the conclusion of the Mean Value

Theorem.

First Derivative Test

- What if they are positive on both sides of the point in question?
- This is called aninflection point

Domain Not a Closed Interval

Example: Find the absolute extrema of

Notice that the interval is not closed. Look graphically:

Absolute Max.

(3, 1)

Optimization Problems

- Identify the unknown(s). Draw and label a diagram as needed.

- Identify the objective function. The quantity to be minimized or maximized.

- Identify the constraints.

4. State the optimization problem.

5. Eliminate extra variables.

6. Find the absolute maximum (minimum) of the objective function.

Optimization - Examples

An open box is formed by cutting identical squares from each corner of a 4 in. by 4 in. sheet of paper. Find the dimensions of the box that will yield the maximum volume.

x

4 – 2x

x

x

x

4 – 2x

The dimensions are 8/3 in. by 8/3 in. by 2/3 in. giving a maximum box volume of V 4.74 in3.

Optimization - Examples

An metal can with volume 60 in3 is to be constructed in the shape of a right circular cylinder. If the cost of the material for the side is $0.05/in.2 and the cost of the material for the top and bottom is $0.03/in.2 Find the dimensions of the can that will minimize the cost.

top and bottom

cost

side

Sub. in for h

Graph of cost function to verify absolute minimum:

2.5

So with a radius ≈ 2.52 in. and height ≈ 3.02 in. the cost is minimized at ≈ $3.58.

Sketch 4 graphs a)1 decreasing and concave upb)1 increasing and concave up, c)1 decreasing and concave down,d)1 increasing and concave down

y

y

x

x

b

a

y

y

x

x

d

c

Download Presentation

Connecting to Server..