Download presentation

Presentation is loading. Please wait.

Published bySimone Linton Modified over 2 years ago

1

2
Rate of Change And Limits What is Calculus? Click to see more.

3
Two Basic Problems of Calculus 1. Find the slope of the curve y = f (x) at the point (x, f (x)) (x, f(x))

4
Area 2. Find the area of the region bounded above by the curve y = f(x), below by the x-axis and by the vertical lines x = a and x = b ab y = f(x) x

5
From BC (before calculus) We can calculate the slope of a line given two points Calculate the slope of the line between the given point P (.5,.5) and another point on the curve, say Q(.1,.99). The line is called a secant line..

6
Slope of Secant line PQ xf(x) Point Q P(0.5, 0.5) Let x values get closer and closer to.5. Determine f(x) values.

7
Slope of Secant line PQ As Q gets closer to P, the Slope of the secant line PQ Gets closer and closer to the slope Of the line tangent to the Curve at P.

8
Figure 1.4: The tangent line at point P has the same steepness (slope) that the curve has at P. Slope of a curve at a point The slope of the curve at a point P is defined to be the slope of the line that is tangent to the curve at point P. In the figure the point is P(0.5, 0.5)

9
In calculus we learn how to calculate the slope at a given point P. The strategy is to take use secant lines with a second point Q. and find the slope of the secant line. Continue by choosing second points Q that are closer and closer to the given point P and see if the difference quotient gets closer to some fixed value.. Slope formula

10
A Find the slope of y = x 2 at the point (1,1) Find the equation of the tangent line. Slope

11
Find slope of tangent line on f(x) =x 2 at the point (1,1) xf(x)Slope of secant between (1,1) and (x, f(x)) Approaching x = 1 from the right Slope appears to be getting close to 2.

12
Find slope of tangent line on f(x) =x 2 at the point (1,1) xf(x)Slope of secant between (1,1) and (x, f(x)) Approaching x = 1 from the left Slope appears to be getting close to 2.

13
Write the equation of tangent line zAs the x value of the second point gets closer and closer to 1, the slope gets closer and closer to 2. We say the limit of the slopes of the secant is 2. This is the slope of the tangent line. zTo write the equation of the tangent line use the point-slope formula

14
Average rate of change (from bc) Find the average velocity if f (t) = 2 + cost on [0, ] f( ) = 2 + cos ( ) = 2 – 1 = 1 f(0) = 2 + cos (0) = = 3 1. Calculate the function value (position) at each endpoint of the interval The average velocity on on [0, ] is If f(t) represents the position of an object as a function of time, then the rate of change is the velocity of the object. 2. Use the slope formula

15
Instantaneous rate of change To calculate the instantaneous rate of change of we could not use the slope formula since we do not have two points. To approximate instantaneous calculate the average rates of change in shorter and shorter intervals to approximate the instantaneous rate of change.

16
To understand the instantaneous rate of change (slope) problem and the area problem, you will need to learn about limits 2.2

17
Limits We write this as: The answer can be found graphically, numerically and analytically. What happens to the value of f (x) when the value of x gets closer and closer and closer (but not necessarily equal) to 2?

18
Graphical Analysis f (x)(x) x What happens to f(x) as x gets closer to 2?

19
Numerical Analysis Start to the left of 2 and choose x values getting closer and closer (but not equal) to 2 x f (x) Use one sided limits Could x get closer to 2? Does f(x) appear to get closer to a fixed number?

20
Numerical Analysis If the limit exists, f(x) must approach the same value from both directions. Does the limit exist? Guess what it is. Start to the right of 2 and choose x values getting closer and closer (but not equal ) to 2 x f (x)

21
Figure 1.8: The functions in Example 7. Limits that do not exist In order for a limit to exist, the function must approach the same value From the left and from the right.

22
Infinite Limits What happens to the function value as x gets closer and closer to 3 from the right? The function increases without bound so we say There is a vertical asymptote at x = 3. x y

23
The line x=a is a Vertical Asymptote if at least one is true. Identify any vertical asymptotes:

24
Graph of f(x) (a)x = 2 is in the domain of f True or false (b) (c)

25
2.3 Functions That Agree at All But One Point If f(x) = g(x) for all x in an open interval except x = c then: Evaluate by direct substitution 2-5 = -3 then Example As x gets closer and closer and closer to 2, the function value gets closer and closer to -3.

26
Analytic Using direct substitution, As x gets closer and closer to 2 (but not equal to 2) f(x) gets closer and closer to 12 = =

27
Compute some limits

28
Basic Limits If b and c are real numbers and is n a positive integer 1. Ex: 2. Ex: 3. Ex: = -2 = 5 = 9 Guess an answer and click to check.

29
Multiplication by a constant b Limit of a sum or difference Limit of a product Limit of a power Limit of a quotient when denominator is not 0. Properties of Limits

30
Properties allow evaluation of limits by direct substitution for many functions. Ex.: As x gets closer and closer to 3, the function value gets closer and closer to 9. Using Properties of Limits

31
Analytic Techniques Direct substitution First substitute the value of x being approached into the function f(x). If this is a real number then the limit is that number. zIf the function is piecewise defined, you must perform the substitution from both sides of x. The limit exists if both sides yield the same value. If different values are produced, we say the limit does not exist.

32
Analytic Techniques Rewrite algebraically if direct substitution produces an indeterminate form such as 0/0 zFactor and reduce zRationalize a numerator or denominator zSimplify a complex fraction When you rewrite you are often producing another function that agrees with the original in all but one point. When this happens the limits at that point are equal.

33
Find the indicated limit = - 5 direct substitution fails Rewrite and cancel now use direct sub. 0 0

34
Find the indicated limit direct substitution fails Rewrite and cancel now use direct sub. 0 0

35
Find the indicated limit calculate one sided limits 7 5 Since the one-sided limits are not equal, we say the limit does not exist. There will be a jump in the graph at x =2

36
Figure 1.24: The graph of f ( ) = (sin )/ . Determine the limit on y = sin θ/θ as θ approaches 0. Although the function is not defined at θ =0, the limit as θ 0 is 1.

37
Figure 1.37: The graph of y = e 1/x for x < 0 shows lim x 0 – e 1/x = 0. (Example 11) A one-sided limit

38
Limits that are infinite (y increases without bound) An infinite limit will exist as x approaches a finite value when direct substitution produces If an infinite limit occurs at x = c we have a vertical asymptote with the equation x = c.

39
Figure 1.50: The function in (a) is continuous at x = 0; the functions in (b) through ( f ) are not. 2.5 Continuity in (a) at x = 0 but not in other graphs.

40
Conditions for continuity A function y = f(x) is continuous at x = c if and only if: The function is defined at x = c The limit as x approaches c exists The value of the function and the value of the limit are equal.

41
Find the reasons for discontinuity in b, c, d, e and f.

42
Figure 1.53: Composites of continuous functions are continuous. Composite Functions Example: is continuous for all reals. If two functions are continuous at x = c then their composition will be continuous.

43
Exploring Continuity Are there values of c and m that make the function continuous At x = 1? Find c and m or tell why they do not exist.

44
Exploring Continuity

45
2.6 Slope of secant line and slope of tangent line

46
s(t) = 8(t 3 – 6 t 2 +12t) 1. Draw a graph. 3. What is the average velocity for the following intervals a. [0, 2], b. [.5, 1.5] c. [.9,1.1] 2. Does the car ever stop? 4. Estimate the instantaneous velocity at t = 1 Position of a car at t hours. t s

47
s(t) = 8(t 3 – 6 t 2 +12t) 3. What is the average velocity for [0, 2], [.5, 1.5] [.9,1.1] 2. Appears to stop at t =2. (Velocity= 0) ts(t) a) 31 mph b) 26 mph c) mph

48
Find an equation of the tangent line to y = 2x 3 – 4 at the point P(2, 12) So, m = 24. Use the point slope form to write the equation

49
Figure 1.62: The tangent slope is f (x 0 + h) – f (x 0 ) h h0h0 lim Slope of the tangent line at x= a f(a+h) – f(a) a a + h P(a, f(a)) Q(a + h, f (a + h))

50
Other form for Slope of secant line of tangent line Let h = x - aThen x = a + h

51
Find an equation of the tangent line at (3, ½) to At a = 3, m = - 1/8 Using the point-slope formula:

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google