How do you remove modulus and rewrite a function in the given interval?

Question

So this thing has been bothering me for a while. I am able to solve basic questions involving absolute values but when a function is defined over an interval, I am mostly stuck on rewriting it.

For example $f(x)= |1-4x^2| \;\; \forall \;\; x \,\epsilon \,[0,1)$

In the texbook, It's rewritten as

$f(x)=1-4x^2 \;\; \forall \;\;x\,\epsilon\,[0,1/2)$

& $\;4x^2-1 \;\; \forall \;\;x\,\epsilon\,[1/2,1)$

I am unable to figure out how this has been done. Any help would be appreciated. Thanks.

Answer 1

Note that the roots of the equation: $f(x) = 1-4x^2$ are: $\frac{1}{2}$ and $-\frac{1}{2}$ (how?). Thus, the behaviour of the absolute value of $f(x)$ depends on the value of $x$ in these ranges.

Note, that as $1-4x^2=(1-2x)(1+2x)$, thus, we can divide our analysis into two cases: (as we are not concerned about the behaviour of $f(x)$ less than $-\frac{1}{2}<0$ because it is not in our range)

• When $0 \leq x < \frac{1}{2}, (1-2x)>0 \implies (1-4x^2)>0$. So, ?

• When $\frac{1}{2} \leq x < 1, (1-2x)<0 \implies (1-4x^2)<0$. So, ?

Answer 2

When $1-4x^2 \geq 0$, then you can get rid of the absolute value. This happens when $0 \leq x \leq 1/2$. If $1-4x^2 < 0$ then you change signs and get rid of the absolute value. Now guess what the range of $x$ is in this case.