The inequality which implies f(x) > 0 - Spivak's Calculus

1. Aug 25, 2013

Seydlitz

Hello,

if you guys would turn to page 117 in Spivak's Calculus, there is the proof for theorem 3. At the last line he stated that this last inequality $|f(x)-f(a)|<f(a)$ implies $f(x)>0$. How can you check this fact?

Can we assume first that $f(x)-f(a)<0$ to eliminate the absolute value, which leads to the inequality $f(x)>0$?

On the second case if $f(x)-f(a)>0$, we have $2f(a)-f(x)>0$, how can we infer that $f(x)>0$

Is it sufficient to show that $f(x)>0$ with the first case? (Logically this means $(f(x)>0)\wedge (f(x)<2f(a))$

Thank You

2. Aug 25, 2013

eigenperson

You have to prove the result for both cases of the sign of $f(x) - f(a)$. If you only prove it when the quantity is negative, then you haven't proved the result. You've only proved it when it is negative.

Fortunately, the other case isn't too hard. I can't figure out how to give a proper hint for this question so I'll just tell you the answer. If $f(x) - f(a) \ge 0$ then $f(x) \ge f(a)$. But an assumption at the beginning of the theorem was that $f(a) > 0$, so you're done.

3. Aug 25, 2013

Mathitalian

First attempt:

From $f(x)-f(a)>0$ we have $f(x)>f(a)$
From $f(x)-f(a)<f(a)$ we have $f(x)<2f(a)$

so $f(a)<f(x)<2f(a)$

but $f(a)<2f(a)\iff f(a)>0\implies f(x)>f(a)>0\implies f(x)>0$

Hope this is helpful (and correct xD)

: I don't have the book

4. Aug 25, 2013

Seydlitz

Ok thanks you all, I understand it now!