Question

Let f : R → R and let f (x)0 for all x ∈ R. Show that f (x) is strictly increasing if and only if the function g(x) = 1/f (x) is strictly decreasing.

Answer 1

To show that a function f(x) is strictly increasing, we need to prove that for any two points a and b in the domain of f, where a < b, f(a) < f(b). If f(x) > 0 for all x in the domain, then the reciprocal of f(x), g(x) = 1/f(x), is defined for all x in the domain. To show that g(x) is strictly decreasing, we need to prove that for any two points a and b in the domain of g, where a < b, g(a) > g(b).

Step-by-step explanation:

To show that a function f(x) is strictly increasing, we need to prove that for any two points a and b in the domain of f, where a < b, f(a) < f(b).

If f(x) > 0 for all x in the domain, then the reciprocal of f(x), g(x) = 1/f(x), is defined for all x in the domain.

To show that g(x) is strictly decreasing, we need to prove that for any two points a and b in the domain of g, where a < b, g(a) > g(b).

Since g(x) = 1/f(x), g(a) > g(b) is equivalent to 1/f(a) > 1/f(b), which can be simplified to f(b) > f(a). This proves that if f(x) is strictly increasing, then g(x) = 1/f(x) is strictly decreasing, and vice versa.