MHB How to Define a Piecewise Function Without Absolute Value Bars?

[paulmdrdo1]

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]

Re: define the function...

I would use the definition:

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

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

 

[paulmdrdo1]

Re: define the function...

I would use the definition:

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

So, what you want to find, is where the function $$f(x)=x^2-1$$ 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]

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:

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

 

[paulmdrdo1]

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:

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

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]

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.

 

[paulmdrdo1]

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]

Re: define the function...

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