Let f:[0,1]-->R be a continuous function such that f(x)>0 for all x contained in [0,1].Show that....?

Question

the reciprocal 1/f:[0,1]-->R is bounded.

Answer 1

The reciprocal function could be unbounded iff there is some point at which the function value is zero. The given conditions preclude the existence of any such point.

Answer 2

strengthen f(x) by employing defining h(x) on [0,a million] as = 0 if x = 0 or a million, and f(x) in any different case. h(x) is non-supply up on [0,a million]. because of the fact that [0,a million] is compact (closed and bounded), so is a twin of [0,a million] under h, so it has a max and a min. If the max and min are the two 0, then h and f are the two consistent = 0, so that's a max and min completed by employing f. If the max or min at the instant are not 0, the max or min occurs in (0,a million), so this is completed by employing f. enable g = -x(x-a million). That has zeroes at 0 and a million, so no minimum in (0,a million), and a max in between at a million/2.

Answer 3

Let m = min[f(x)] on [0,1]. This means that every other value in f(x) >= m.

From this, every value in 1/f(x) <= 1/m = M. Therefore, 1/f(x) has a range of [0,M], meaning that 1/f(x) is bounded.