The Intermediate Value Theorem If a function f (x) is continuous in the closed interval [ a,b] then [ ]

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1 Lecture 2 5B Evaluating Limits Limits x ---> a The Intermediate Value Theorem If a function f (x) is continuous in the closed interval [ a,b] then [ ] the y values f (x) must take on every value on the interval f (a), f (b) Note: This theorem guarantees that as the values of x take on every value on the closed interval [ a, b ] the values of y will take on every value on the interval [ f (a), f (b) ]. This does not it the y values from being smaller than f(a) or larger than f(b) somewhere on the open interval (a, b ). It does guarantee that as the values of x take on every values on [ a,b ] the values of y will take on every [ ] value on the interval f (a), f (b) Limits Many its questioned can be answered based on what it means for a function f (x) to be continuous for all values of x = a in the interval (b < a <c). If f (x) is continuous in (b < a <c) then the point ( a,f(a) ) exists for all vales of a. This means that If f (x) is continuous in (b < a <c) the values of x approach a from both sides of c and the y values of f (x) must approach f(a). f(a) (a, f(a) ) x = b x = a x = c Lecture 2 5B Page 1 of Eitel

2 Theorem 1 If f (x) is continuous in the open interval b < a < c ( ) then f (a) R then there is a point on the graph of f (x) at ( a, f (a)) and x a f (x) = f (a) which implies x a f (x) = f (a) and x a + f (x) = f (a) x a and f (x) = f (a) x a f(x) = f(a) The it of f(x) as x approaches a from the LEFT is f(a) This theorem includes the following its x a + f(x) = f(a) The it of f(x) as x approaches a from the RIGHT is f(a) x a f(x) = f(a) The it of f(x) as x approaches a is f(a) another wording of this states that If f (x) is continuous in ( b < a < c) then f(a) will be a real number. and f(a) = L then x a f(x) = L and x a + f(x) = L x a and f(x) = L Lecture 2 5B Page 2 of Eitel

3 f (x) = 4x 7 find Example 1 x 4x 7 f (x) = 4x 7 is continuous in the neighborhood on both sides of x = so theorem 1 can be used. 1. x 4x 7 2. x + 4x 7. x 4x 7 = f () = 4 = 5 ( ) 7 = f () = 4 = 5 ( ) 7 = f () = 4 = 5 ( ) 7 The its above taken together say that as x approaches the left side of the graph of f (x) = 4x 7 and the right side of the graph of f (x) = 4x 7 both approach the point (, 5). f (x) = x 2 +1 find Example 2 x 2 x2 +1 f (x) = x 2 +1 is continuous in the neighborhood on both sides of x = 2 so theorem 1 can be used. 1. x 2 x x 2 + x2 +1. x 2 x2 +1 = f(2) = f(2) = f(2) = 2 = 1 ( ) 2 +1 = 2 = 1 ( ) 2 +1 = ( 2) 2 +1 = 1 The its above taken together say that as x approaches the left side of the graph of f(x) = x 2 +1and the right side of the graph of f(x) = x 2 +1both approach the point (2,1). Lecture 2 5B Page of Eitel

4 Example f (x) = x 2 x 1 find x 4 x 2 x 1 The denominator of f (x) equals 0 when x = 1 so f (x) = x 2 x 1 f (x) = x 2 x 1 but we are asking for the it in the neighborhood of x = 4 is not continuous at x = 1 is continuous in the neighborhood on both sides of x = 4 so Theorem 1 can be used. x 4 x 2 x 1 = f(4) = = 2 Note: You only need to find f(4) to know that as x approaches 4 the left side of the graph of and the right side of the graph of both approach the point (4, 2/ ). Example 4 f (x) = x 2 x 2 4 find x 1 x 2 x 2 4 The denominator of f (x) equals 0 when x = 2 or 2 so f (x) is not continuous at x = 2 or 2 but we are asking for the it in the neighborhood of x = 1 f (x) is continuous in the neighborhood on both sides of x = 4 so theorem 1 can be used. x 1 x 2 x 2 4 = f (1) = = 1 Note: You only need to find f(1) to know that as x approaches 1 the left side of the graph of and the right side of the graph of both approach the point (1, 1/ ). Lecture 2 5B Page 4 of Eitel

5 Example 5 f(x) = x find x 4 x (4, 2 ) f (x) is not defined for x < 0. f (x) is not continuous at x = 0 but we are asking for the it in the neighborhood of x = 4 f (x) is continuous in the neighborhood on both sides of x = 4 so Theorem 1 can be used. x x 4 = f (4) = 4 = 2 Note: You only need to find f(4) to know that as x approaches 4 the left side of the graph of and the right side of the graph of both approach the point (4, 2). Lecture 2 5B Page 5 of Eitel

6 Where does Theorem 1 fail End Points and Theorem 1 x 0 x and Find x 0 + x and x 0 x Theorem 1 requires that f (x) be continuous in the neighborhood around x = 0 x is not defined for x < o so f (x) is NOT continuous in the neighborhood around x = 0 Theorem 1cannot be used to find the it as x > 0 However the it questions can be answered by examining the graph of the function. The graph has an end point at x = 0 and the endpoint has coordinates (0,0) x is continuous in the neighborhood to the right of x = 0 As x approaches 0 from the right the values of y approach 0 so x 0 + x = 0 x is NOT continuous in the neighborhood to the left of x = 0 so x 0 x = DNE and since the left and right sided its are not equal x 0 x = DNE Lecture 2 5B Page 6 of Eitel

7 Where does Theorem 1 fail Vertical Asymptotes. y x 2 f(x) = (x 2) find x 2 (x 2) Theorem 1 requires that f (x) be continuous in the neighborhood around x = 2 f(2) is not defined so f(x) is NOT continuous in the neighborhood around x = 2 so Theorem 1 CANNOT be used. However the it question can be answered, but not using Theorem 1. If there is a vertical asymptote at x = a then the left and right sided its involve ±. We have already examined how to find the its in the neighborhood around x = a where x = a is a vertical asymptote. test f(2.1) = f(1.9) = + = test f(2 +.1) = f(2.1) = + + = + f(1.9) < 0 so x 2 (x 2) = f(2.1) > 0 so x 2 + (x 2)2 = + x 2 (x 2) = DNE Lecture 2 5B Page 7 of Eitel

8 Where does Theorem 1 fail Hole f(x) = (x +1) ( x 2 ) (x 2) find x 2 f(x) Theorem 1 requires that f (x) be continuous in the neighborhood around x = 2 f(2) is not defined so f(x) os NOT continuous in the neighborhood around x = 2 so Theorem 1 CANNOT be used. However the it question can be answered, but not using Theorem 1. There is a hole in the graph of f(x) at (2,). As x approaches 2 from both side of 2 the y values approach the y coordinate of the hole which is. The it as x approaches 2 is. x 2 (x +1) ( x 2) = (x 2) Note: Most functions will be continuous in the interval x 1 < a < x 2 1 can be used. f(x) = f(a). x a ( ) for most values of a and Theorem For most functions we will encounter, Theorem 1 fails for very few x values. However most of the it questions you will be asked to solve will involve finding its for values of x where Theorem cannot be used. We will need to develop other Theorems to deal with this situations. Lecture 2 5B Page 8 of Eitel