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

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1 Lecture 4-6B1 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) Theorem 1 If f (x) is continuous in the open interval b < a < c ( ) then f (a) R and there is 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) Lecture 4 6B1 Page 1 of Eitel

2 f (x) x 7 find Example 1 x 4x 7 f (x) x 7 is continuous in the neighborhood on both sides of x so theorem 1 can be used. x 4x 7 2. x + 4x 7. x 4x 7 The its above taken together also say that as x approaches the left side of the graph of f (x) x 7 and the right side of the graph of f (x) x 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 so theorem 1 can be used. x 2 x x 2 + x2 +1. x 2 x2 +1 ( ) 2 +1 ( ) 2 +1 = ( 2) 2 +1 The its shown above taken together say that f(x) = x 2 +1is continuous in the neighborhood around x. The its above taken together also say that as x approaches 2 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 (2,5). Lecture 4 6B1 Page 2 of Eitel

3 Example find The denominator of f (x) equals 0 when x = 1 so is not continuous at x = 1 but we are asking for the it in the neighborhood of x is continuous in the neighborhood on both sides of x so theorem 1 can be used Example 4 x 2 4 find x 1 x 2 4 The denominator of f (x) equals 0 when x or 2 so f (x) is not continuous at x 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 so theorem 1 can be used. x 1 x Lecture 4 6B1 Page of Eitel

4 Example 5 f(x) = x find x The domain of f (x) is x > 0. f (x) The graph has an end point at x = 0 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 f (x) is continuous in the neighborhood on both sides of x so theorem 1 can be used. x 2. + x. x x a f (x) = f (a) requires that f (x) is continuous in the neighborhood of x = a. If there is a hole, point or vertical asymptote at x = a Theorem 1 cannot be used to find the it as x approaches a form both sides. There may be an answer to the it question but Theorem 1 cannot be used to find the it. If there is a point at a, f (a) then x a + f (x) = If there is a point at then x a 1 f (x) = ( ) and the graph is continuous in the neighborhood to the right of of x = a f (a) ( a, f (a)) and the graph is continuous in the neighborhood to the left of of x = a f (a) If there is a vertical asymptote at x = a then the left and right sided its involve ±. Theorem 1 cannot be used to find the it. We have already examined how to find the its in the neighborhood of vertical asymptotes. Example 6 f (x) = () () find x 2 f (x) f (x) is not defined for x. f (x) is not continuous at x f (x) is NOT continuous in the neighborhood on both sides of x so Theorem 1 CANNOT be used. Lecture 4 6B1 Page 4 of Eitel

5 If there is a hole in the graph at x = a then f (a) does not exist. There may be an answer to the it question but Theorem 1 cannot be used to find it. This case will be examined in the next portion of the lecture. Example 7 f (x) = ()() () find x 2 f (x) f (x) is not defined for x. f (x) is not continuous at x f (x) is NOT continuous in the neighborhood on both sides of x so Theorem 1 CANNOT be used. NOTE: Do not factor and cancel out like terms in the function before determining the domain restrictions. Lecture 4 6B1 Page 5 of Eitel