Limits and Continuity AP Calculus AB Foerster: Exploration

Limits and Continuity AP Calculus AB Foerster: Exploration

Limits and Continuity AP Calculus AB Foerster: Exploration 2-1a Objective: Find the limit of a function that approaches an indeterminate form at a particular value of x and relate it to the definition.

1. Plot on your grapher the graph of this function. Use a friendly window with x = 3 as a grid point, but with the grid turned off. Sketch the results here. Show the behavior of the function in a neighborhood of x = 3.

1. Plot on your grapher the graph of this function. Use a friendly window with x = 3 as a grid point, but with the grid turned off. Sketch the results here. Show the behavior of the function in a neighborhood of x = 3.

2. Substitute 3 for x in the equation for f(x). What form does the answer take? What name is given to an expression of this form? 3. The graph of f has a removable discontinuity at x = 3. The y-value at this discontinuity is the limit of f(x) as x approaches 3. What number does this limit equal?

2. Substitute 3 for x in the equation for f(x). What form does the answer take? What name is given to an expression of this form? The result is called an indeterminate form. 3. The graph of f has a removable discontinuity at x = 3. The y-value at this discontinuity is the limit of f(x) as x approaches 3. What number does this limit equal?

The limit is y = 2 as x approaches 3. It appears that the hole can be filled with the point (3, 2). 4. Make a table of values of f(x) for each 0.1 unit change in x-value from 2.5 through 3.5. 5. Between what two numbers does f(x) stay when x is kept in the open interval (2.5, 3.5)?

6. Simplify the fraction for f(x) and evaluate this simplified function for x = 3. 6. Simplify the fraction for f(x) and evaluate this simplified function for x = 3.

2.2 Limits: A Numerical and Graphical Approach The goal in this section is to define limits and study them using numerical and graphical techniques. We begin with the following question: How do the values of a function f(x) behave when x approaches a number c, whether or not f(c) is defined?

To explore this question, well experiment with the function Evaluate f(0)

To explore this question, well experiment with the function This is called an indeterminate form We can compute f(x) for values of x close to 0. To describe the trend, we use the phrase x approaches 0 to

indicate that x takes on values (both positive and negative) that get closer and closer to 0. The notation for this is , and more specifically we write if x approaches 0 from the right

if x approaches 0 from the left Complete the following table to determine the value f(x) approaches as x approaches 0. x x

x x 1 -1 1 -1 0.5 -0.5

0.5 -0.5 0.1 -0.1 0.1 -0.1 0.05 -0.05

0.05 -0.05 0.01 -0.01 0.01 -0.01 0.005 -0.005

0.005 -0.005 0.001 -0.001 0.001 -0.001 Complete the following table to determine the value f(x) approaches

as x approaches 0. x 1 0.5 0.1 0.05 0.01 0.01

0.005 0.005 0.001 0.001 x 0.841470985 0.958851077

0.998334166 0.999583385 0.999983333 0.999983333 0.999995833 0.999995833 0.999999833 0.999999833

-1 -0.5 -0.1 -0.05 -0.01 -0.01 -0.005

-0.005 -0.001 -0.001 0.841470985 0.958851077 0.998334166 0.999583385

0.999983333 0.999983333 0.999995833 0.999995833 0.999999833 0.999999833

The conclusion found from the table of values is supported by the graph of f(x). The point (0, 1) is missing from the graph because f(x) is not defined at x = 0 However, the graph approaches this missing point as x approaches 0 from the left and right.

We say that the limit of f(x) as x 0 is equal to 1 0 is equal to 1 Investigate the following limit by completing the table 8.9

9.1 8.99 9.01 8.999

9.001 8.9999 9.0001

Investigate the following limit by completing the table 8.9 5.983286 9.1

6.016621 8.99 5.998332 9.01

6.001666 8.999 5.999833 9.001

6.000167 8.9999 5.999833 9.0001

6.000017 Caution: Numerical investigations are often suggestive, but may be misleading in some cases. Using tables and looking at graphs can give us a good idea of what is

happening and what it appears as though the limit is approaching. However, they should not be used to determine an exact limit In this unit we will discover methods to determine exact limits. The limits discussed so far are two=sided

To show that , we should check that f(x) converges to L as x approaches c through values both larger and smaller than c That is, we must show In some instances, f(x) may approach L from one side of c without necessarily approaching it from the other side, or f(x) may be defined on only ones side of c. The limit itself exists if both one-sided limits exist and are equal.

Investigate the one-sided limits of as . Does exist?

Investigate the one-sided limits of as . Does exist? The function f(x) in the figure below is not defined at c = 0, 2, 4. Investigate the one- and two-sided limits at these points.

Some functions f(x) tend to or - as x approaches a value c In these cases, does not exist, but we say that f(x) has an infinite limit. if f(x) increases without bound as x 0 is equal to 1 c if f(x) decreases without bound as x 0 is equal to 1 c

Keep in mind that and - are not numbers and the above notation means that the limit does not exist because the graph increases (or decreases) without bound When f(x) approaches or - as x approaches c from one or both sides, the line x = c is called a vertical asymptote.

Investigate the limits graphically: Investigate the limits graphically:

Investigate the limits graphically: Investigate the limits graphically:

Investigate the limits graphically: Summary If f(x) approaches a limit as x 0 is equal to 1 c, then the limit value L is unique. A limit does not exist if:

: The one-sided limits are not equal : The graph increases or decreases without bound f(x) oscillates The limit may exist even if f(c) is not defined. If a one- or two-sided limit is , the vertical line x = c is called a vertical asymptote.

Warm-Up! Evaluate the limit 2.3 Basic Limit Laws In the previous section we relied on graphical and numerical approaches to investigate limits and estimate their values. In the next sections we go beyond this intuitive approach and

develop tools for computing limits in a precise way. Basic Limit Laws If and exist, then Sum and Difference Law: exists and Constant Multiple Law: For any number k, exists and

Product Law: exists and Basic Limit Laws Continued If and exist, then Quotient Law: If , then exists and Powers and Roots: If p, q are integers with q 0, then exists and

Theorem For any constants k and c, Some limits can be evaluated by a simple substitution.

Substitution is valid when the function is continuous, a concept we shall study later. Try direct substitution as the first step in evaluating limits. Examples

Examples Warm-up Sketch the graph of the function.

1. 2. 3. Which graph appears to be continuous? 2.4 Limits and Continuity On a white board, take one minute to write down what you think a formal definition for a function to be continuous could be. Include

what it would mean for a function to be discontinuous. Take one minute to share with a partner. Each pair will take one minute to share with the class. In everyday speech, the word continuous means having no breaks or interruptions. In calculus, continuity is used to describe functions whose graphs have no breaks.

Look at graph A below. Determine f(2) and . Look at graph B below. Determine f(2) and .

Look at graph C below. Determine f(2) and . Look at graph D below. Determine f(2) and .

Look at graph E below. Determine f(2) and . Look at graph F below. Determine f(2) and .

Complete the table from the information found. Graph A B C D E

F f(2) Continuous? Complete the table from the information found. Graph

f(2) Continuous? A 6

DNE No B 4

DNE No C 4

4 Yes D 6

4 No E DNE

DNE No F DNE

4 No There are three conditions that must hold for a function to be continuous. Based on the table above, what do you think those three conditions are?

Definition Continuity at a Point Assume that f(x) is defined on an open interval containing x = c. Then f is continuous at x = c if If the limit does not exist, or if it exists but is not equal to f(c), we say

that f has a discontinuity (or is discontinuous) at x = c. A function f(x) may be continuous at some points and discontinuous at others. If f(x) is continuous at all points in an interval I, then f(x) is said to be continuous on I. If f(x) is continuous at all points in its domain, then f(x) is simply called continuous.

Keep in mind that continuity at a point x = c requires more than just the existence of a limit. Three conditions must hold: 1. f(c) is defined

2. exists 3. They are equal If exists but is not equal to f(c), we say that f has a removable discontinuity at x = c. We say that f(x) has a nonremovable discontinuity at x = c if one or both of the one-sided limits is infinite.

Theorem Continuity of the Inverse Function If f(x) is continuous on an interval I with range R, and if exists, then is continuous with domain R. Theorem Continuity of Composite Functions If g is continuous at x = c, and f is continuous at x = g(c), then the composite function F(x) = f(g(x)) is continuous at x = c.

Summary Definition: f(x) is continuous at x = c if If does not exist, or if it exists but does not equal f(c), then f is discontinuous at x = c If f(x) is continuous at all points in its domain, f is simply called

continuous Three common types of discontinuities: Removable discontinuity: the limit exists but does not equal f(c) Jump discontinuity: the one-sided limits both exist but are not equal Infinite discontinuity: the limit is infinite as x approaches c 2.5 Evaluating Limits Algebraically

Substitution can be used to evaluate limits when the function in question is known to be continuous. When we study derivatives, we will be faced with limits where f(c) is not defined. Many of these limits can be evaluated if we use algebra to rewrite the formula for f(x). Our strategy when direct substitution gives us an indeterminate form, is to transform f(x) algebraically into a new expression that is

defined and continuous at x = c, and then evaluate the limit by substitution. Examples

Theorem Important Trigonometric Limits Examples Limits Review

Complete the table and use the result to estimate the limit. 1. 2.

x f(x) 1.9 1.99

1.999 2.001 2.01 2.1

x f(x) -0.1 -0.01 -0.001

0.001 0.01 0.1

Complete the table and use the result to estimate the limit. 1. 2. x

f(x) 1.9 1.99 1.999 2.001 2.01 2.1 .34483 .33445 .33344 .33322 .33223 .32258

x f(x) -0.1 .29112 -0.01

.28892 -0.001 .28869 0.001 .28865

0.01 .28843 0.1 .28631

3. Use the graph to find the limit (if it exists). If the limit does not exist, explain why. 3.

Use the graph to find the limit (if it exists). If the limit does not exist, explain why. Use the graph to find the limit (if it exists). If the limit does not exist, explain why.

3. For , find Use the graph to find the limit (if it exists). If the limit does not exist, explain why.

3. For , find State the three types of behavior associated with nonexistence of a limit and draw a function to represent each. State the three types of behavior associated with nonexistence of a

limit and draw a function to represent each. : The one-sided limits do not approach the same number. : The graph increases or decreases without bound. The graph oscillates as a.

b. c. d. Sketch a graph of a function f that satisfies the given values. is undefined

a. b. c. d. Sketch a graph of a function f that satisfies the given values. is undefined This says there is a hole when x = 0

This clarifies that the hole is at (0, 4) and that the graph should approach 4 from both sides. This says there is a point This says there is a that the graph should both sides.

at (2, 6) hole at (2, 3) but approach 3 from Find the limit, if it exists.

7. 7. 8. 8. 9. 9.

Find the limit, if it exists. 10. 10. 11. 11.

12. 12. indeterminate form; try a different method; Find the limit, if it exists

13. 13. indeterminate form; try a different method; 14. 14. indeterminate form; try a different method;

Find the limit, if it exists. 15. 15. indeterminate form; try a different method; 16. 16.

Graph the function and determine if the function is continuous, using correct math notation, at x = 1. 17. Because ,

Since , f(x) is continuous at x = 1. Graph the function and determine if the function is continuous, using correct math notation, at x = 1. 18.

Since , does not exist and therefore f(x) is not continuous at x = 1. Find the value of c that makes the function continuous. 19.

Find the value of c that makes the function continuous. 20. Limits and Continuity Quiz

Exploration 2-4a: Continuous and Discontinuous Functions Let f be the piecewise function defined by where k stands for a constant.

1. Plot the graph of f for k = 1. Sketch the result. 2. Function f is discontinuous at x = 2. Tell what it means for a function to be discontinuous. Let f be the piecewise function defined by

where k stands for a constant. 1. Plot the graph of f for k = 1. Sketch the result. Let f be the piecewise function defined by where k stands for a constant.

2. Function f is discontinuous at x = 2. Tell what it means for a function to be discontinuous. f(a) does not exist, or does not exist, or

3. Find and . (The second limit will be in terms of k.) what must be true of these two limits for f to be continuous at x = 2? 4. Find the value of k that makes f continuous at x = 2. Sketch the graph of f for this value of k.

3. Find and . (The second limit will be in terms of k.) what must be true of these two limits for f to be continuous at x = 2? 4. Find the value of k that makes f continuous at x = 2. Sketch the graph of f for this value of k. 5. The graph in Problem 4 has a cusp at x = 2. What is the origin of the word cusp, and why is it appropriate to use in this context?

6. Suppose someone asks, Is f(x) increasing or decreasing at x = 2 with k as in Problem 4? How would you have to answer that question? What, then, can you conclude about the derivative of a function at a point where the graph has a cusp? 5. The graph in Problem 4 has a cusp at x = 2. What is the origin of the word cusp, and why is it appropriate to use in this context? 6. Suppose someone asks, Is f(x) increasing or decreasing at x = 2

with k as in Problem 4? How would you have to answer that question? What, then, can you conclude about the derivative of a function at a point where the graph has a cusp? Differentiability

Defintion: A function f is differentiable at a if So, how can a function fail to be differentiable? When the limit does not exist. So, how can a limit fail to exist? The right and left hand limits do not agree or the limit is infinity

So, the slope to the right and left do not agree. This is called a cusp. It is like a corner or kink in the graph. Or the slope is infinite or there is a vertical tangent line.

Differentiability is connected to local linearity. If f is differentiable at x = a, then when we zoom in at the point (a, f(a)), the graph straightens out and appears more and more like a line. At a cusp, no amount of zooming will straighten out the graph. At a discontinuity, no amount of zooming will straighten out the

graph even though the left and right rates of change may be the same. Refresher Statement: If p, then q. Converse: If q, then p. Inverse: If not p, then not q. Contrapositive: If not q, then not p.

If the statement is true, the contrapositive must be true. If the converse is true, the inverse must be true. Differentiability Implies Continuity If f is differentiable at x = a, then f is continuous at x = a. What is the contrapositive to this statement? If f is not continuous at x = a, then f is not differentiable at x = a. Both of these statements are true.

Differentiability Implies Continuity If f is differentiable at x = a, then f is continuous at x = a. What is the converse statement? Is it true? Why or why not? If f is continuous at x = a, then f is differentiable at x = a. What is the inverse statement? Is it true? Why or why not? If f is not differentiable at x = a, then f is not continuous at x = a.

But, if a function f is continuous at x = a, then f is not necessarily differentiable at x = a Or, if a function f is not differentiable at x = a, then f may be continuous at x = a. Therefore, the converse and inverse statements are not true. Differentiability and Continuity Practice

1. State whether the graph illustrates a function that i. has left and right limits at the marked value of x. ii. has a limit at the marked value of x. iii. is continuous at the marked value of x. If it is not continuous there, explain why.

iv. is differentiable at the marked value of x. If it is not differentiable there, explain why. 1. State whether the graph illustrates a function that i. has left and right limits at the marked value of x.

ii. has a limit at the marked value of x. iii. is continuous at the marked value of x. If it is not continuous there, explain why. iv. is differentiable at the marked value of x. If it is not differentiable there, explain why. 1. State whether the graph illustrates a function that i.

has left and right limits at the marked value of x. ii. has a limit at the marked value of x. iii. is continuous at the marked value of x. If it is not continuous there, explain why. iv. is differentiable at the marked value of x. If it is not differentiable there, explain why.

1. State whether the graph illustrates a function that i. has left and right limits at the marked value of x. ii. has a limit at the marked value of x. iii. is continuous at the marked value of x. If it is not continuous there,

explain why. iv. is differentiable at the marked value of x. If it is not differentiable there, explain why. 1. State whether the graph illustrates a function that i. has left and right limits at the marked value of x.

ii. has a limit at the marked value of x. iii. is continuous at the marked value of x. If it is not continuous there, explain why. iv. is differentiable at the marked value of x. If it is not differentiable there, explain why. 1. State whether the graph illustrates a function that

i. has left and right limits at the marked value of x. ii. has a limit at the marked value of x. iii. is continuous at the marked value of x. If it is not continuous there, explain why. iv. is differentiable at the marked value of x. If it is not differentiable

there, explain why. 1. State whether the graph illustrates a function that i. has left and right limits at the marked value of x. ii. has a limit at the marked value of x.

iii. is continuous at the marked value of x. If it is not continuous there, explain why. iv. is differentiable at the marked value of x. If it is not differentiable there, explain why. 1. State whether the graph illustrates a function that i.

has left and right limits at the marked value of x. ii. has a limit at the marked value of x. iii. is continuous at the marked value of x. If it is not continuous there, explain why. iv. is differentiable at the marked value of x. If it is not differentiable there, explain why.

For the graphs below, list the x-values for which the function appears to be (i) not continuous and (ii) not differentiable. Give a reason for your answer. For the graphs below, list the x-values for which the function appears to be (i) not continuous and (ii) not differentiable. Give a reason for your answer.

For the functions graphed below, does the function appear to be differentiable on the interval of x-values shown. For the functions graphed below, does the function appear to be differentiable on the interval of x-values shown. Explain what is wrong with each of the following statements. A function f that is not differentiable at x = 0 has a graph with a

sharp corner at x = 0. Besides a sharp corner, the function could be not continuous or have a vertical tangent line. Explain what is wrong with each of the following statements. If f is not differentiable at a point then it is not continuous at that point. A function can be continuous but just have a sharp corner at a point,

like f(x) = |x| at x = 0. Give an example graphically or algebraically of a continuous function that is not differentiable at x = 2. Explain your reasoning. One example: f(x) = |x 2| Give an example graphically or algebraically of a function not continuous at x = 2 that is differentiable at x = 2. Explain your

reasoning. Not possible since if a function is not continuous it cannot be differentiable. Are the statements below true or false? If a statement is true, give an example illustrating it. If a statement is false, give a counterexample. There is a function which is continuous on [1, 5] but not

differentiable at x = 3. True. Sample answer: f(x) = |x 3| Are the statements below true or false? If a statement is true, give an example illustrating it. If a statement is false, give a counterexample. If a function is differentiable, then it is continuous. True.

Are the statements below true or false? If a statement is true, give an example illustrating it. If a statement is false, give a counterexample. If a function is continuous, then it is differentiable. False. Sample answer: f(x) = |x 3| Are the statements below true or false? If a statement is true, give

an example illustrating it. If a statement is false, give a counterexample. If a function is not continuous, then it is not differentiable. True Are the statements below true or false? If a statement is true, give an example illustrating it. If a statement is false, give a counterexample.

If a function is not differentiable then it is not continuous. False. Sample answer: f(x) = |x 3| Exploration: Intermediate Value Theorem 1. The figure shows the graph of a function f

defined on the closed interval [1, 5]. Why does the function f not meet the hypotheses of the intermediate value theorem? 2. For function f in problem 1, explain why the conclusion of the intermediate value theorem is false if y = 5? 3. For function f in problem 1, is there a value of x for which y = 3? Does this fact contradict the intermediate value theorem?

1. The figure shows the graph of a function f defined on the closed interval [1, 5]. Why does the function f not meet the hypotheses of the intermediate value theorem? The hypotheses states that f must be continuous and at x = 3, the function f has a jump discontinuity. 1. For function f in problem 1, explain why the

conclusion of the intermediate value theorem is false if y = 5? f(1) = 2 and f(5) = 8 y = 5 is between f(1) and f(5). But the conclusion that there exists at least one number c such that f(c) = 5 is false. Looking at the graph there are no values of x where y = 5. 1. For function f in problem 1, is there a value

x for which y = 3? Does this fact the intermediate value theorem? of contradict Yes there is a value, x = 2, where f(2) = 3. This does not contradict the intermediate value theorem because the theorem only guarantees a

value of c when the function is continuous, it does not rule out there might be a value of c such that f(c) = 3. 4. The figure shows a function g defined the closed interval [0, 5]. Does the meet the hypotheses of the theorem on that

on function g intermediate value interval? Why or why not? 5. Is there a value of x = c in [0, 5] for every value of k between g(0) and g(5) for which g(c) = k? Does this fact contradict the intermediate value theorem?

6. How does the function g in problem 4 show the intermediate value theorem is not an if and only if theorem? 4. The figure shows a function g defined on the closed interval [0, 5]. Does the function g meet the hypotheses of the intermediate value theorem on that interval? Why or why not?

The function g does not meet the hypotheses of the intermediate value theorem on the interval [0. 5] because g is not continuous at x = 3. 4. Is there a value of x = c in [0, 5] for every value of k between g(0) and g(5) for which g(c) = k? Does this fact contradict the intermediate value

theorem? Yes there is a value of x = c in [0, 5] for every value of k between g(0) and g(5) for which g(c) = k. This does not contradict the theorem because the theorem does not rule out the possibility that the conclusion could be true. 4. How does the function g in problem 4 show the intermediate value theorem

is not an if and only if theorem? if and only if means the converse of the intermediate value theorem is true. But this example shows that even though there is a value of x = c in [0, 5] for every value of k between g(0) and g(5) for which g(c) = k, the function g is not continuous. Intermediate Value Theorem

The Intermediate Value Theorem (IVT) says, roughly speaking, that a continuous function cannot skip values. Consider a plane that takes off and climbs from 0 to 10,000 meters in 20 minutes. The plan must reach every altitude between 0 and 10,000 meters during this 20-minute interval. Thus, at some moment, the planes altitude must have been exactly 8371 meters. Of course, this assumes that the planes motion is continuous, so its

altitude cannot jump abruptly from, say, 8000 to 9000 meters. To state this conclusion formally, let A(t) be the planes altitude at time t. The IVT asserts that for every altitude M between 0 and 10,000, there is a time t0 between 0 and 20 such that A(t0) = M. In other words, the graph of A(t) must intersect the horizontal

line y = M. By contrast, a discontinuous function can skip values. Intermediate Value Theorem If f(x) is continuous on a closed interval [a, b] and f(a) f(b),

then for every value M between f(a) and f(b), there exists at least one value such that f(c) = M. Example Prove that the equation sin x = 0.3 has at least one solution in the

interval . The IVT can be used to show the existence of zeros of functions. If f(x) is continuous and takes on both positive and negative values say f(a) < 0 and f(b) > 0 then the IVT guarantees that f(c) = 0 for some c between a and b. Corollary to the IVT If f(x) is continuous on [a, b] and if f(a) and f(b) are nonzero and have opposite signs, then f(x) has a zero in

(a, b). We can locate zeros of functions to arbitrary accuracy using the Bisection Method. Example

Show that has a zero in (0, 2). Then locate the zero more accurately using the Bisection Method. Exploration: Mean Value Theorem For the function f graphed below, there is a value of x = c between 3 and 7 at which the tangent line to the graph of f is parallel to the secant line through (3, f(3)) and (7, f(7)).

a. Draw the secant line and the tangent line. b. From the graph, c __________ c. Is f differentiable on (3, 7)? __________ d. Is f continuous on [3, 7]? __________ For the function f graphed below, there is a value of x = c between 3 and 7 at which the tangent line to the graph of f is parallel to the secant line through (3, f(3)) and (7, f(7)).

a. Draw the secant line and the tangent line. b. From the graph, c 5.5 c. Is f differentiable on (3, 7)? Yes d. Is f continuous on [3, 7]? Yes The function f in problem 1 has two values of x = c between x = 1

and x = 7 at which equals the slope of the secant line through (1, f(1)) and (7, f(7)). That is, the tangent line is parallel to the secant line. a. Draw the secant line and the tangent lines. b. From the graph, c __________ and c __________ c. Is f differentiable on (1, 7)? d. Is f continuous on [1, 7]?

The function f in problem 1 has two values of x = c between x = 1 and x = 7 at which equals the slope of the secant line through (1, f(1)) and (7, f(7)). That is, the tangent line is parallel to the secant line.

a. Draw the secant line and the tangent lines. b. From the graph, c 5.6 and c 2.3 c. Is f differentiable on (1, 7)? Yes d. Is f continuous on [1, 7]? Yes For the function g graphed below:

a. Draw a secant line through (1, g(1)) and (5, g(5)). b. Is g differentiable on (1, 5)? c. Is g continuous on [1, 5]? d. Tell why there is no value of x = c between x = 1 and x = 5 at which equals the slope of the secant line draw in part (a).

For the function g graphed below: a. Draw a secant line through (1, g(1)) and (5, g(5)). b. Is g differentiable on (1, 5)? No c. Is g continuous on [1, 5]? Yes

d. Tell why there is no value of x = c between x = 1 and x = 5 at which equals the slope of the secant line draw in part (a). The slope of the secant line is about and along the left branch at x = 1, increasing so there is no point left branch where . On

rate of change is negative. and is along the the right branch the

The function g from problem 3 does have a value x = c in (1, 4) for which equals the slope of the secant line through (1, g(1)) and (4, g(4)). a. Draw the secant line and the tangent line below. b. From the graph, c __________. c. Is g differentiable on (1, 4)? d. Is g continuous on [1, 4]?

The function g from problem 3 does have a value x = c in (1, 4) for which equals the slope of the secant line through (1, g(1)) and (4, g(4)). a. Draw the secant line and the tangent line below.

b. From the graph, c 3. c. Is g differentiable on (1, 4)? Yes d. Is g continuous on [1, 4]? Yes For the function h graphed below:

a. Draw a secant line through (5, h(5)) and (7, h(7)). b. Is h differentiable on (5, 7)? c. Is h continuous on [5, 7]? d. Tell why there is no value of x = c in (5, 7) at which equals the slope of the secant line drawn in part (a).

For the function h graphed below: a. Draw a secant line through (5, h(5)) and (7, h(7)). b. Is h differentiable on (5, 7)? Yes c. Is h continuous on [5, 7]? No d. Tell why there is no value of x = c in (5, 7) at which equals the slope of the secant line drawn in part (a). The slope of the secant line is about -1. The

values of are all positive on (5, 7). The graph below is the function h from problem 5. a. Draw a secant line through (5, h(5)) and (9, h(9)). b. Is h differentiable on (5, 9)?

c. Is h continuous on [5, 9]? d. There is a value of x = c in (5, 9) where equals the slope of the secant line drawn in part (a). Draw the tangent line. Estimate the value of c. c __________ The graph below is the function h from problem 5.

a. Draw a secant line through (5, h(5)) and (9, h(9)). b. Is h differentiable on (5, 9)? Yes c. Is h continuous on [5, 9]? No d. There is a value of x = c in (5, 9) where equals the slope of the secant line drawn in part (a). Draw the tangent line. Estimate the value of c. c 6

The graph below is the function h from problem 5. a. Draw a secant line through (0, h(0)) and (9, h(9)). b. Show that there are two points x = c in (0, 9) where equals the slope of the secant line drawn in part (a). Draw the tangent lines and estimate the values of c. c __________ and c __________

c. Is h differentiable on (0, 9)? d. Is h continuous on [0, 9]? The graph below is the function h from problem 5. a. Draw a secant line through (0, h(0)) and (9, h(9)).

b. Show that there are two points x = c in (0, 9) where equals the slope of the secant line drawn in part (a). Draw the tangent lines and estimate the values of c. c 3 and c 5.5 c. Is h differentiable on (0, 9)? No d. Is h continuous on [0, 9]? No

The Mean Value Theorem states: If f is differentiable on (a, b) and f is continuous on [a, b], then there is a number c (the mean value) in (a, b) such that . The slope of the tangent line equals the slope of the secant line. For which problem(s) are: a. the hypotheses and conclusion true? b. the hypotheses and conclusion not true?

c. the conclusion is true but not the hypotheses? The Mean Value Theorem states: If f is differentiable on (a, b) and f is continuous on [a, b], then there is a number c (the mean value) in (a, b) such that . The slope of the tangent line equals the slope of the secant line.

For which problem(s) are: a. the hypotheses and conclusion true? 1, 2, and 4 b. the hypotheses and conclusion not true? 3 and 5 c. the conclusion is true but not the hypotheses? 6 and 7 Summarize what you learned about the mean value theorem.

Summarize what you learned about the mean value theorem. One of the big ideas is that the hypotheses guarantee the conclusion that instantaneous rate of change must equal the average rate of change, but the conclusion can be true even when the hypotheses is not true. IVT and MVT Review

The IVT can be used to show the existence of zeros of functions. If f(x) is continuous and takes on both positive and negative values say f(a) < 0 and f(b) > 0 then the IVT guarantees that f(c) = 0 for some c between a and b. Corollary to the IVT If f(x) is continuous on [a, b] and if f(a) and f(b) are nonzero and have opposite signs, then f(x) has a zero in

(a, b). We can locate zeros of functions to arbitrary accuracy using the Bisection Method. The Mean Value Theorem states: If f is differentiable on (a, b) and f is continuous on [a, b], then there is a number c (the mean value) in

(a, b) such that . The slope of the tangent line equals the slope of the secant line. For which problem(s) are: a. the hypotheses and conclusion true? b. the hypotheses and conclusion not true? c. the conclusion is true but not the hypotheses?

1. Use the IVT to show that takes on the value 9 for some x in [1,2 ]. 2. Show that has a zero in [0, 1]. Apply the Bisection Method twice to narrow the window. 3. Find a point c satisfying the conclusion of the MVT for on the interval [-4, 5].

IVT and MVT Quiz Fill in the blank. 1. IVT: If f is ____________________ on a closed interval [a, b] and k is a number between _____ and _____ then there is at least one number c

in [a, b] such that f(c) = k. 2. Corollary to the IVT: If f(x) is ____________________ on [a, b] and if f(a) and f(b) have ________________ signs, then f(x) has a __________ in (a, b). 3. MVT: If f is ____________________ on (a, b) and f is ____________________ on [a, b], then there is a number c (the mean value) in (a, b) such that .

4. Use the IVT to show that takes on the value 6 for some x in [2, 5]. 5. Show that has a zero in [1, 2]. Apply the Bisection Method twice to narrow the window. 6. Find a point c satisfying the conclusion of the MVT for on the interval [9, 25]. Limits at Infinity

We have considered limits as x approaches a number c. It is also important to consider limits where x approaches or -, which we refer to as limits at infinity. In applications, limits at infinity arise naturally when we describe the long-term behavior of a system as shown in the graph.

The notation x 0 is equal to 1 indicates that x increases without bound, and x 0 is equal to 1 - indicates that x decreases (through negative values) without bound. if f(x) gets closer and closer to L as x 0 is equal to 1 if f(x) gets closer and closer to L as x 0 is equal to 1 - In either case, the line y = L is called a horizontal asymptote. Infinite limits describe the asymptotic behavior of a function,

which is the behavior of the graph as we move out to the right or left. Example Discuss the asymptotic behavior in the figure below.

Example Discuss the asymptotic behavior in the figure below. There is a horizontal asymptote of y = 7 as we move to the right. There is a horizontal asymptote of y = 3 as we move to the left.

A function may approach an infinite limit as x 0 is equal to 1 . Similar notation is used if f(x) approaches -

The graph of is shown. What are the following limits? The graph of is shown. What are the following limits?

Limits at infinity do not always exist. For example, oscillates indefinitely so do not exist. The limits at infinity of the power functions are easily determined

Theorem For all n > 0, If n is a whole number, Theorem

The asymptotic behavior of a rational function depends only on the leading terms of its numerator and denominator. If , then Proof of this theorem is in the textbook.

1. 2. 3. 4. Use the theorem to evaluate the following limits.

Given the graph of behaves as the graph of : If n = m, If n < m, If n > m, Think about the end behaviors of the graph of g(x) to determine if it is or

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