1. 1 Example 1 Explain why the Intermediate Value Theorem does or does not apply to each of the following functions. (a) f(x) = 1/x with domain [1,2]. Solution The function f is continuous on the interval [1,2]. Hence the Intermediate Value Theorem applies to f. It says that if A = f(a) = 1/a and B = f(b) = 1/b are selected with 1  a  b  2, then all numbers between A and B are in the range of f. That is, if B = 1/b  C  1/a = A then C = f(1/C). Moreover, since 1  a  1/C  b  2, the number 1/C is in the domain [1,2] of f.

2. 2 (b) g(x) = 1/x with domain (0,1]. Solution The function g is continuous on the interval (0,1]. Hence the Intermediate Value Theorem applies to g. It says that if A = g(a) = 1/a and B = g(b) = 1/b are selected with 0 < a  b  1, then all numbers between A and B are in the range of g. That is, if B = 1/b  C  1/a = A then C = g(1/C). Moreover, since 0 < a  1/C  b  1, the number 1/C is in the domain (0,1] of g.

3. 3 (c) h(x) = 1/x with domain (1,  ). Solution The function h is continuous on the interval (1,  ). Hence the Intermediate Value Theorem applies to h. It says that if A = h(a) = 1/a and B = h(b) = 1/b are selected with 1 < a  b, then all numbers between A and B are in the range of h. That is, if B = 1/b  C  1/a = A then C = h(1/C). Moreover, since 1 < a  1/C  b, the number 1/C is in the domain (1,  ) of h.

4. 4 (d) k(x) = 1/x Solution The domain of k consists of all numbers for which the denominator x of 1/x is nonzero, i.e all numbers other than zero. Thus the domain of k is (- ,0)  (0,  ) which is not an interval. Hence the Intermediate Value Theorem does not apply to k. Note that the conclusion of the Intermediate Value Theorem is not valid for k: -1 = k(-1) < 0 < k(1) = 1 and –1, 1 are in the domain of k. However, 0 is not in the range of k - there is no number whose reciprocal equals zero. We can also see this from the graph of k below. Although the graph of k has height –1 at x=-1 and height +1 at x=+1, it never has height 0. The reason is apparent – the graph of k is not continuous at x=0 where it has a vertical asymptote.