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Define the function...

paulmdrdo

Active member
May 13, 2013
386
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

Administrator
Staff member
Feb 24, 2012
13,775
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
May 13, 2013
386
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.
i have two answers

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

Administrator
Staff member
Feb 24, 2012
13,775
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
May 13, 2013
386
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

Administrator
Staff member
Feb 24, 2012
13,775
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
May 13, 2013
386
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.
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: define the function...

Yes, that would be correct, albeit not the most efficient piecewise statement of the function though.