1 / 22

Drill: Find f(2)

Drill: Find f(2). f(x) = f(x) = f(x) = f(x) =. Lesson 2.1 Rates of Changes and Limits. day # 1 homework: p. 66: 1-6, 8-14(even), 16-18, 20-28 (even) day #2 homework: p. 66-67: 29-50. Average and Instantaneous Speed.

cecily
Download Presentation

Drill: Find f(2)

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. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Drill: Find f(2) • f(x) = • f(x) = • f(x) = • f(x) =

  2. Lesson 2.1Rates of Changes and Limits day # 1 homework: p. 66: 1-6, 8-14(even), 16-18, 20-28 (even) day #2 homework: p. 66-67: 29-50

  3. Average and Instantaneous Speed • A moving body’s average speed during an interval of time is found by dividing the distance covered by elapsed time. • Example 1: • A rock breaks lose from the top of a tall cliff. What is its average speed during the first 2 seconds of fall? • Solution: • Use the equation y(t) = 16t2, where y = distance and t = time in seconds. • For the first two seconds of the fall, starting time is 0 and ending time is 2. • Use the formula

  4. Finding Instantaneous Speed • To find instantaneous speed at a time t and some time h seconds from t, use the formula: • Find the speed of the rock from example 1 at the instant t = 2 seconds. • If we are looking for the speed to be instantaneous, h = 0, but that can only be evaluate at the end, so 64 + 16(0) = 64 ft/sec.

  5. Definition of Limit • The limit is a method of evaluating an expression as an argument approaches a value. This value can be any point on the number line and often limits are evaluated as an argument approaches infinity or minus infinity. The following expression states that as x approaches the value c the function approaches the value L. • When looking at a graph, the limit of a rational function (f(x)/g(x)) is the HORIZONTAL asymptote!

  6. Determining the limit by substitution and confirm graphically (put in calc)

  7. Determine the limit graphically. Then prove algebraically!(Always FACTOR with rational expressions!) the limit appears to be 1/2

  8. Determine the limit graphically. Then prove algebraically!(Always FACTOR with rational expressions!) the limit appears to be -1/2

  9. Properties of Limits • Known limits: • limit of the function with the constant k: • limit of the identity function of x = c: • Properties of Limits: If L, M, c and k are real number and

  10. Properties of Limits • Sum Rule • The limit of the sum of 2 functions is the sum of their limits • Difference Rule • The limit of the difference of 2 functions is the difference of their limits • Product Rule • The limit of a product of 2 functions is the product of their limits

  11. Properties of Limits • Constant Multiple Rule • The limit of a constant times a function is the constant times the limit of the function • Quotient Rule • The limit of a quotient of two functions is the quotient of their limits, provided that the denominator is not zero. • Power Rule • If r and s are integers and s ≠ 0, then

  12. Using Properties of Limits • Use the observations that and , and the properties of limits to evaluate the following: • limxc (x4 + 4x2 -3) • limxc x4 + limxc 4x2 – limxc 3 • c4 + 4c2 -3 • limxcx4 + x2 -1 x2 + 5 • limxc x4 + limxc x2 - limxc 1limxc x2 +limxc 5 • c4 + c2 -1 c2 + 5

  13. Drill • Evaluate the following limits • limx3 x2 (2 – x) • limx2x2 + 2x + 4 x + 2 • limx0tanx *you will want to use the tanx = sinx/cosx x • Why does the following limit NOT exist? • limx2x3 – 1 x - 2

  14. Drill solutions • By substitution • -9 • 3 • Using the product rule: • limx0 (sinx/x) · (1/cosx) • limx0 (sinx/x*) · limx0(1/cosx) *Look at graph • 1·(1/cosx) • 1 · 1 = 1 • since you cannot factor the given rational function, you must just substitute 2; however, by doing so, the denominator would be 0, which is not possible.

  15. One-sided and Two-sided Limits • Sometimes the values of a function f tend to different limits as x approaches a number c from the left and from the right. • Right-hand limit: • The limit of f as x approaches c from the right. • Left-hand limit: • The limit of f as x approaches c from the left • Example: Use the graph to the right

  16. Definition of Step Function • A step function is a special type of function whose graph is a series of line segments. • The graph of a step function looks like a series of small steps. • There are two types of step functions: rounding UP and rounding DOWN. • Rounding UP • Greatest Integer Function • f(x) = -int (-x) • Ceiling Function • Example f(.4) = 1

  17. Rounding Down • Least Integer Function • f(x) = int(x) • Floor Function • Example: f(2.1) = 2

  18. One-Sided and Two-sided Limits • A function f(x) has a limit as x approaches c if and only if (IFF) the right-hand and left-hand limits at c exist and are equal.

  19. Exploring Right-handed and Left-handed Limits 1 0 1 • Use the graph and function below to determine the limits. Therefore, no limit exists as x1, since they are NOT the same The rest of the limits regarding this problem can be found on page 64.

  20. Exploring Left-handed and Right-Handed Limits 1 2 2 3

  21. Sandwich Theorem • If g(x) < f(x) < h(x) for all x ≠ c in some interval about c, and limxc g(x) = limxch(x) = L, then limxcf(x) = L • Using the sandwich theorem: • Show that the limx0 [x2 sin(1/x)] = 0 • We know that all of the values of the sine function lie between -1 and 1 (range) • The greatest value that sin(1/x) can have is 1 and the lowest value is -1, so… • -x2<x2 sin(1/x)] < x2 • Because the limx0-x2 = 0 and limx0x2= 0, • then limx0 [x2 sin(1/x)] = 0

  22. Closure

More Related