Question

by removing the absolute values, express (4-x^2)/|x+2| as a peicewise-defined function

Answer 1

\[f(x) = |x+2| = \left\{\begin{array}{rcc}
-x-2 & \text{if} & x <2\\
x+2& \text{if} & x>2
\end{array}
\right.
\]is a start

Answer 2

damn all that formatting and it is wrong!

Answer 3

the restrictions wouldn't be x< -2 and x> -2

Answer 4

\[f(x) = |x+2| = \left\{\begin{array}{rcc}
-x-2 & \text{if} & x <-2\\
x+2& \text{if} & x>-2
\end{array}
\right.\]

Answer 5

better?

Answer 6

yes but the answer choices are x + 2, x > 2, −(x + 2), x <2
2. f(x) = x−2, x > −2, 2−x, x < −2.
3. f(x) = 2−x, x > −2, x−2, x < −2.
4. f(x) = x + 2, x > −2, −(x + 2), x < −2.
5. f(x) = x−2, x > 2, 2−x, x < 2.
6. f(x) = −(x + 2), x > 2, x + 2, x < 2.

Answer 7

fine i just did the denominator

Answer 8

factor the numerator
if \(x>-2\) there will be a common factor of \(x+2\) top and bottom
cancel it

Answer 9

if \(x<-2\) there will be a factor of \(x+2\) in the top and \(-x-2\) in the bottom and
\[\frac{x+2}{-x-2}=-1\]

Answer 10

I did the top which would simplify It -(x-2)(x+2)/(x+2) so the then the x+2 cancels out

Answer 11

im confused as to what your saying

Answer 12

top is \((2-x)(2+x)\) right?

Answer 13

if \(x>-2\) you can cancel the \(2+x\) top and bottom and get just \(2-x\)

Answer 14

whereas if \(x<-2\) the denominator is \(-x-2\)

Answer 15

you do not have a factor of \(-x-2\) in the numerator, however you can still cancel because
\[\frac{2+x}{-x-2}=-1\] leaving you with \(-(2-x)\) or \(x-2\)

Answer 16

I get it so (2-x) x>-2
-x-2 x<-2