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How do you write #f(x) = 3 - |2x + 3|# into piecewise functions?

1 Answer
Dec 27, 2017

Answer:

#f(x) = -2x# for #x>=-3/2# and #2x+6# for #x<-3/2#.

Explanation:

It's really the absolute value you need to worry about.

If you're rewriting absolute values of the form #|u|# start with finding the zero of #u#: #2x+3=0\rightarrow x=-3/2#.

Now figure out where #2x+3# is positive and negative.

By inspection (or picking values to substitute) #2x+3<0# for #x<-3/2# and #2x+3>=0# for #x>=-3/2#.

In general #|u| = u# when #u>=0# and #|u|=-u# when #u<0#, so now we have:

#|2x+3| = 2x+3# for #x>=-3/2# and #|2x+3|=-2x-3# for #x<-3/2#. We make those substitutions.

For #x>=-3/2#: #f(x) = 3- (2x+3) = -2x#
For #x<-3/2#: #f(x) = 3 - (-2x-3) =2x+6#

So that's our function. (Not sure how I'd write a piecewise function on here using the normal notation.)

#f(x) = -2x# for #x>=-3/2# and #2x+6# for #x<-3/2#.