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Published byDwayne Haseltine Modified about 1 year ago

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1 Example 3 (a) Find the range of f(x) = 1/x with domain [1, ). Solution The function f is decreasing on the interval [1, ) from its largest value f(1) = 1 towards its horizontal asymptote y = 0. Hence the range of f is contained in the interval (0,1]. Since the function f is continuous on the interval [1, ), the Intermediate Value Theorem applies to f. Hence the range of f is an interval from a to 1. Since the values of f are all positive, a 0. Since the x-axis is a horizontal asymptote of f, there are large numbers x for which f(x) a are arbitrarily close to zero. Hence a = 0, and the range of f is the interval (0,1].

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2 (b) Find the range of g(x) = 1/x with domain (0,1]. Solution The function g is decreasing on the interval (0,1] from its vertical asymptote x = 0 towards its smallest value g(1) = 1. Hence the range of g is contained in the interval [1, ). Since the function g is continuous on the interval (0,1], the Intermediate Value Theorem applies to g. Hence the range of g is an interval from 1 to b. Since the y-axis is a vertical asymptote of g there are numbers x close to 0 for which g(x) b are arbitrarily large. Hence b = , and the range of g is the interval [1, ).

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3 (c) Find the range of h(x) = 1/x with domain (0, ). Solution The function h is decreasing on the interval (0, ) from its vertical asymptote x = 0 towards its horizontal asymptote y = 0. Since the values of h are all positive, the range of h is contained in the interval (0, ). The function h is continuous on the interval (0, ), the Intermediate Value Theorem applies to h. Hence the range of h is an interval from a to b. Since the y-axis is a vertical asymptote of h there are numbers x close to 0 for which h(x) b are arbitrarily large. Hence b = . Since the x-axis is a horizontal asymptote of h there are large numbers x for which h(x) a are arbitrarily close to zero. Hence a=0 and the range of h is the interval (0, ).

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