- #1

CuriousBanker

- 190

- 24

Theorem: Suppose f is continuous at a, and f(a)>0. Then there is a number @>0 such that f(x)>0 for all x satisfying |x-a|<@

Proof: Consider the case f(a)>0 Since f is continuous at a, if &>0 there is a @>0 such that, for all x, if |x-a|<@, then |f(x)-f(a)<&

Since f(a)>0, we can take f(a) as & Thus there is @>0 so that for all x, if |x-a|<@ then |f(x)-f(a)|<f(a)

So far, I understand all of this. The part I don't understand is the very last line of the proof, which says:

and this last inequality implies f(x)>0

I don't get how if |x-a|<@ then |f(x)-f(a)|<f(a) implies that f(x)>0.

INTUITIVELY I get it...because if |f(x)-f(a)|<f(a) that means the distance between f(x)-f(a) is smaller than the number f(a), and f(a)=f(a)-0, so if f(x) were negative, the distance between f(x) and f(a) would be greater than the distance between f(a) and 0. I get it...but I jus't cant prove it with rigor.

Thank you in advance for your help

Somebody posted this answer on yahoo answers:

If |f(x) - f(a)| < f(a), then -f(a) < f(x) - f(a) < f(a), so in particular -f(a) < f(x) - f(a). Adding f(a) to both sides of this inequality gives 0 < f(x).

But I dont see why the absolute value signs got dropped