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#### paulmdrdo

##### Active member

- May 13, 2013

- 386

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!

- Thread starter paulmdrdo
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- Thread starter
- #1

- May 13, 2013

- 386

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!

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- #2

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.

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- May 13, 2013

- 386

i have two answersI 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?

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- #4

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:

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

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- May 13, 2013

- 386

i based my first answer on this definition of abs. value 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:

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

|x|= { x, if x>0

.......{ 0, if x=0

.......{ -x, if x<0

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- #7

- May 13, 2013

- 386

oh i see. but what if i add that x>1 to my first answer?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.

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.

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