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

[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

 

[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.

 

[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)

[Edit]: I don't have the book

 

[Seydlitz]

Ok thanks you all, I understand it now!