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

Answer 1

Eva Abramson

Use the piecewise definition of the absolute value function.

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

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):}#

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Answer 2

Ethan Adkins

The piecewise function for ( y = |4x + 1| + 2x - 3 ) would be:

[ y = \begin{cases} (4x + 1) + 2x - 3 & \text{if } 4x + 1 \geq 0 \ -(4x + 1) + 2x - 3 & \text{if } 4x + 1 < 0 \end{cases} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.