1

Example 2 (a) Find the range of f(x) = x2+1 with domain [0,2].

Solution The function f is increasing on the interval [0,2] from its smallest value f(0) = 1 to its largest value f(2) = 5. Hence the range of f is contained in the interval [1,5]. Since the function f is continuous on the interval [0,2], the Intermediate Value Theorem applies to f. It says that every number between f(0) = 1 and f(2) = 5 is in the range of f. That is, the range of f contains the interval [1,5]. Thus the range of f is the interval [1,5].

y

x2

1

5

y = x2+1

2

(b) Find the range of g(x) = x2+1 with domain (0,2].

Solution The function g is continuous on the interval (0,2]. Hence the Intermediate Value Theorem applies to g, and the range of g is an interval. Since g is the restriction of f obtained by deleting the number 0 from the domain of f, the range of g is the range [1,5] of f with the number f(0) = 1 deleted. Thus the range of g is the interval (1,5].

y

x2

1

5

y = x2+1

3

(c) Find the range of h(x) = x2+1 with domain (0,2).

Solution The function h is continuous on the interval (0,2). Hence the Intermediate Value Theorem applies to h and the range of h is an interval. Since h is the restriction of f obtained by deleting the numbers 0, 2 from the domain of f, the range of h is the range [1,5] of f with the numbers f(0) = 1 and f(2) = 5 deleted. Thus the range of h is the interval (1,5).

y

x2

1

5

y = x2+1