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. y

x1 a 1/C b 2

1/2B

A

1

C

y= 1/x

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.

y

xa 1/C b 1

1B

A

C

y= 1/x

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.

y

x1 a 1/C b

B

A

1

C

y= 1/x

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. y

x-11

y= 1/x

1-1