Calculus - inequality problem. I have this inequality :
$$|g(x)-B|<\frac{|B|}{2}$$ 
$$-\frac{|B|}{2}<g(x)-B<\frac{|B|}{2}$$
$$B-\frac{|B|}{2}<g(x)<B+\frac{|B|}{2}$$
I don't understand how it possible to conclude from the inequality that
$$\frac{|B|}{2}<|g(x)|$$ 
Thanks.
 A: $$|B-\frac{|B|}{2}|<|g(x)|$$
Case 1: $$B\geq0$$
$$|B-\frac{|B|}{2}|=\frac{B}{2}=\frac{|B|}{2}<|g(x)|$$
Case 2: $$B<0$$
$$|B-\frac{|B|}{2}|=\frac{-3B}{2}=\frac{3|B|}{2}<|g(x)| \implies \frac{|B|}{2}<|g(x)|$$
A: Lets start from where you left off:
$B-\frac{|B|}{2}<g(x)<B+\frac{|B|}{2}$
This implies that:
$|B\pm \frac{|B|}{2}|=\{\frac{|B|}{2},\frac{3|B|}{2}\}$
Hence,
$\frac{|B|}{2}<|g(x)|<\frac{3|B|}{2} \rightarrow \frac{|B|}{2}<|g(x)|$
A: suppose ($g(x)\gt0$ and $B\lt0$) or ($g(x)\lt0$ and $B\gt0$)
then $\mid g(x)-B\mid=\mid g(x)\mid + \mid B\mid\gt \mid B\mid\ge\frac{\mid B\mid}{2}$
If $g(x)=0$ then $\mid g(x)-B\mid = \mid B\mid\ge\frac{\mid B\mid}{2}$
If $B=0$ then $\mid g(x)-B\mid = \mid g(x)\mid\ge \frac{\mid B\mid}{2}=0$
now suppose ($g(x)\gt0$ and $B\gt0$), then your inequality becomes
$$B-\frac{|B|}{2}\lt g(x)\lt B+\frac{|B|}{2}$$
$$\frac{B}{2}\lt g(x)\lt \frac{3B}{2}$$
$$\mid\frac{B}{2}\mid\lt \mid g(x)\mid\lt \mid\frac{3B}{2}\mid$$
Finally suppose ($g(x)\lt0$ and $B\lt0$), then your inequality becomes
$$B-\frac{|B|}{2}\lt g(x)\lt B+\frac{|B|}{2}$$
$$\frac{3B}{2}\lt g(x)\lt \frac{B}{2}$$
$$-\frac{3B}{2}\gt -g(x)\gt -\frac{B}{2}$$
$$-\frac{B}{2}\lt -g(x)\lt -\frac{3B}{2}$$
$$\mid\frac{B}{2}\mid\lt \mid g(x)\mid\lt \mid\frac{3B}{2}\mid$$
I think that this establishes that $|g(x)-B|\lt\frac{|B|}{2}\implies \frac{|B|}{2}\lt|g(x)|$
A: If $B>0$ then the first inequality implies $B/2 < g(x)< 3B/2$.
When $B<0$, the first inequality implies $B/2>g(x)>3B/2$
