# Define the function...

#### paulmdrdo

##### Active member
Define the function piecewise without absolute value bars

f(x)=|x^2-1| - what is the core or prevailing function here? is it the absolute value function or the squaring?

please solve this. and show the steps. thanks!

#### MarkFL

Staff member
Re: define the function...

I would use the definition:

$$\displaystyle |x|\equiv\begin{cases}-x & x<0\\ x & 0\le x \\ \end{cases}$$

So, what you want to find, is where the function $$\displaystyle f(x)=x^2-1$$ is negative, and where is it non-negative.

#### paulmdrdo

##### Active member
Re: define the function...

I would use the definition:

$$\displaystyle |x|\equiv\begin{cases}-x & x<0\\ x & 0\le x \\ \end{cases}$$

So, what you want to find, is where the function $$\displaystyle f(x)=x^2-1$$ is negative, and where is it non-negative.

f(x) = {x^2-1 , if x<-1
.........{0 , if x=-1 or x=1
.........{1-x^2, if -1<x<1

or

f(x) = {x^2-1 , if x<=-1
.........{1-x^2, if -1<x<1
.........{x^2-1, if x>=1

are they both correct?

#### MarkFL

Staff member
Re: define the function...

The second one is correct...do you see why the first is not?

The function $f(x)=x^2-1$ is negative on $(-1,1)$, otherwise is is non-negative, so another way we could define $|f(x)|$ is:

$$\displaystyle |f(x)|=\begin{cases}1-x^2 & |x|<1 \\ x^2-1 & 1\le|x| \\ \end{cases}$$

#### paulmdrdo

##### Active member
Re: define the function...

The second one is correct...do you see why the first is not?

The function $f(x)=x^2-1$ is negative on $(-1,1)$, otherwise is is non-negative, so another way we could define $|f(x)|$ is:

$$\displaystyle |f(x)|=\begin{cases}1-x^2 & |x|<1 \\ x^2-1 & 1\le|x| \\ \end{cases}$$
i based my first answer on this definition of abs. value function.

|x|= { x, if x>0
.......{ 0, if x=0
.......{ -x, if x<0

#### MarkFL

Staff member
Re: define the function...

You left out $1\le x$. Also, we really don't see special cases for when the expression is equal to zero, we just need to differentiate between negative and non-negative.

#### paulmdrdo

##### Active member
Re: define the function...

You left out $1\le x$. Also, we really don't see special cases for when the expression is equal to zero, we just need to differentiate between negative and non-negative.
oh i see. but what if i add that x>1 to my first answer?

f(x) = {x^2-1 , if x<-1
.........{0 , if x=-1 or x=1
.........{1-x^2, if -1<x<1
.........{x^2-1, if x>1

would this be correct? I just want to explore on the different possible answers. please bear with me.