How do you write #y = |4x + 1| + 2x - 3# as a piecewise function?

1 Answer
Oct 15, 2017

Answer:

Use the piecewise definition of the absolute value function.

#|f(x)| = {(f(x); f(x)>=0),(-f(x);f(x)<0):}#

Explanation:

Substitute #4x+1# for #f(x)#:

#|4x+1| = {(4x+1; 4x+1 >=0),(-(4x+1);4x+1<0):}#

Simplify the inequalities:

#|4x+1| = {(4x+1; x >=-1/4),(-(4x+1);x<-1/4):}#

Substitute the piecewise parts into the given equation and append the domain restrictions:

#y = {(4x+1 + 2x - 3; x >=-1/4),(-(4x+1) + 2x - 3;x<-1/4):}#

Simplify the pieces:

#y = {(6x- 2; x >=-1/4),(-2x - 4;x<-1/4):}#