anonymous
  • anonymous
by removing the absolute values, express (4-x^2)/|x+2| as a peicewise-defined function
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
\[f(x) = |x+2| = \left\{\begin{array}{rcc} -x-2 & \text{if} & x <2\\ x+2& \text{if} & x>2 \end{array} \right. \]is a start
anonymous
  • anonymous
damn all that formatting and it is wrong!
anonymous
  • anonymous
the restrictions wouldn't be x< -2 and x> -2

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
\[f(x) = |x+2| = \left\{\begin{array}{rcc} -x-2 & \text{if} & x <-2\\ x+2& \text{if} & x>-2 \end{array} \right.\]
anonymous
  • anonymous
better?
anonymous
  • anonymous
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.
anonymous
  • anonymous
fine i just did the denominator
anonymous
  • anonymous
factor the numerator if \(x>-2\) there will be a common factor of \(x+2\) top and bottom cancel it
anonymous
  • anonymous
if \(x<-2\) there will be a factor of \(x+2\) in the top and \(-x-2\) in the bottom and \[\frac{x+2}{-x-2}=-1\]
anonymous
  • anonymous
I did the top which would simplify It -(x-2)(x+2)/(x+2) so the then the x+2 cancels out
anonymous
  • anonymous
im confused as to what your saying
anonymous
  • anonymous
top is \((2-x)(2+x)\) right?
anonymous
  • anonymous
if \(x>-2\) you can cancel the \(2+x\) top and bottom and get just \(2-x\)
anonymous
  • anonymous
whereas if \(x<-2\) the denominator is \(-x-2\)
anonymous
  • anonymous
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 \(x-2\)
anonymous
  • anonymous
I get it so (2-x) x>-2 -x-2 x<-2

Looking for something else?

Not the answer you are looking for? Search for more explanations.