How do you write $\frac{x^{2} - 9}{| x + 3 |}$ $(x^2-9) / ( |x+3| )$ as a piecewise function?

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Answer 1 · 2018-06-18T10:11:49

$f (x) = x - 3$ $f(x)=x-3$ if $x > - 3$ $x > -3$ and $f (x) = - (x - 3)$ $f(x)=-(x-3)$ if $x < - 3$ $x<-3$

Not that

$\frac{x + 3}{| x + 3 |} = 1$ $(x+3)/|x+3|=1$ if $x > - 3$ $x > -3$

$\frac{x + 3}{| x + 3 |} = - 1$ $(x+3)/|x+3|=-1$ if $x < - 3$ $x<-3$

How do you write (x^2-9) / ( |x+3| )x2−9|x+3|(x^2-9) / ( |x+3| ) as a piecewise function?