The first part is true because any continuous map of a compact set is compact.Once f is no longer required to be continuous however, pretty much anything goes. For example, we can have
f(x) := |x−c| for x≠c and 1 for x=c,
where c ∈ [a, b]. Then f will be positive on [a, b] but 1/f will be unbounded near c.
Answers & Comments
Well,
f is continuous on [a, b] with f(x) > 0 for all a ≤ x ≤ b,
therefore
1/f is continuous on [a, b] with 1/f(x) > 0 on the whole interval
hope it' ll help !!
The first part is true because any continuous map of a compact set is compact.Once f is no longer required to be continuous however, pretty much anything goes. For example, we can have
f(x) := |x−c| for x≠c and 1 for x=c,
where c ∈ [a, b]. Then f will be positive on [a, b] but 1/f will be unbounded near c.