Presentation is loading. Please wait.

Presentation is loading. Please wait.

AP Calculus BC Thursday, 03 September 2015 OBJECTIVE TSW (1) determine continuity at a point and continuity on an open interval; (2) determine one- sided.

Similar presentations


Presentation on theme: "AP Calculus BC Thursday, 03 September 2015 OBJECTIVE TSW (1) determine continuity at a point and continuity on an open interval; (2) determine one- sided."— Presentation transcript:

1 AP Calculus BC Thursday, 03 September 2015 OBJECTIVE TSW (1) determine continuity at a point and continuity on an open interval; (2) determine one- sided limits and continuity on a closed interval; (3) use properties of continuity; and (4) understand and use the Intermediate Value Theorem. T-Shirt sales: S – XL$7.00 2XL – 4XL$9.00 –Due by Tuesday, 08 September 2015.

2 Sec. 2.6: Continuity

3 Informally, a function f is continuous at x = c if there is no “interruption” in the graph of f at x = c. That is, the graph of f has no holes, jumps, or gaps at c.

4 Sec. 2.6: Continuity

5

6 Types of discontinuities: HoleGapHole NOTE: The third type of discontinuity is an asymptote.

7 Sec. 2.6: Continuity Ex:Discuss the continuity of the following: f has a domain of (–∞, 0) (0, ∞), so f is continuous at every x-value in its domain. f has a nonremovable discontinuity (asymptote) at x = 0.

8 Sec. 2.6: Continuity Ex:Discuss the continuity of the following: g has a domain of (–∞, 1) (1, ∞), so g is continuous at every x-value in its domain. g has a removable discontinuity (hole) at x = 1.

9 Sec. 2.6: Continuity Ex:Discuss the continuity of the following: h has a domain of (–∞, ∞), so h is continuous at every x-value. h is everywhere continuous.

10 Sec. 2.6: Continuity Ex:Discuss the continuity of the following: y has a domain of (–∞, ∞), so y is continuous at every x-value. y is everywhere continuous

11 Sec. 2.6: Continuity One-Sided Limits A one-sided limit considers only one side of a given value. Ex:The limit from the right means that x approaches c from values greater than c. A similar meaning is given to the limit from the left.

12 Sec. 2.6: Continuity Ex:Find the following:

13 Sec. 2.6: Continuity Ex:Using the definition of continuity, determine if f is continuous at x = 2.

14 Sec. 2.6: Continuity Ex:Using the definition of continuity, determine if f is continuous at x = 2.

15 Sec. 2.6: Continuity The two one-sided limits must equal each other.

16 Sec. 2.6: Continuity

17 Functions that are continuous at every point in their domain: a)Polynomial functions b)Rational functions c)Radical functions d)Trigonometric functions

18 Sec. 2.6: Continuity

19

20 Combining these properties with continuous functions allows you to state that many functions are continuous. Ex:State why the following are continuous: a)f(x) = x 2 + 1 – sin x f is the sum of a polynomial and a trig function

21 Sec. 2.6: Continuity Ex:State why the following are continuous: b) f is a composition of a radical function and a polynomial

22 Sec. 2.6: Continuity Ex:State why the following are continuous: c) f is the product of a polynomial function and a trig function

23 Sec. 2.6: Continuity

24 f is continuous on [a, b].  three c’s  f(c) = k. f is not continuous on [a, b].  no c’s  f(c) = k.

25 Sec. 2.6: Continuity Ex:Use the Intermediate Value Theorem to show that f(x) = x 3 +2x – 1 has at least one zero in [0, 1]. i)f(x) is continuous (because it’s a polynomial) ii)f(0) = –1 f(1) = 2 f(0) < 0 < f(1)  By the IVT,  at least one zero in [0, 1]  f(c) = 0.

26 Sec. 2.6: Continuity Ex:(a) Verify that the Intermediate Value Theorem applies in the indicated interval for f(x). (b) Find the value of c guaranteed by the theorem. f(x) = x 2 – 6x + 8, [0, 3], f(c) = 0 (a)i) f is continuous on [0, 3] (polynomial) ii) f(0) = 8 and f(3) = –1 f(0) > f(c) = 0 > f(3)  By the IVT,  at least one zero in [0, 3]  f(c) = 0.

27 Sec. 2.6: Continuity Ex:(a) Verify that the Intermediate Value Theorem applies in the indicated interval for f(x). (b) Find the value of c guaranteed by the theorem. f(x) = x 2 – 6x + 8, [0, 3], f(c) = 0 (b)Find c. x 2 – 6x + 8 = 0 (x – 4)(x – 2) = 0 x = 4, x = 2 Since x = 2 is in the interval and f(2) = 0, c = 2


Download ppt "AP Calculus BC Thursday, 03 September 2015 OBJECTIVE TSW (1) determine continuity at a point and continuity on an open interval; (2) determine one- sided."

Similar presentations


Ads by Google