How do you write $f (x) = 3 - | 2 x + 3 |$ $f(x) = 3 - |2x + 3|$ as a piecewise function?

Question

1826 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-13T12:12:04

Use the definition of the absolute value function:

$|a| ={(a" for "a>=0),(-a" for "a<0):}$

and then simplify the domain restrictions.

Applying the definition to $|2x+3|$ :

$|2x+3| = {((2x+3)" for "2x+3>=0),(-(2x+3)" for "2x+3<0):}$

Simplify the domain restrictions:

$|2x+3| = {((2x+3)" for "2x>=-3),(-(2x+3)" for "2x<-3):}$

$|2x+3| = {((2x+3)" for "x>=-3/2),(-(2x+3)" for "x<-3/2):}" [1]"$

Now that we have simplified the domain restrictions, we write $f(x)$ with the right side of equation [1] replacing $|2x+3|$ :

$f(x) = {(3 - (2x+3)" for "x>=-3/2),(3 - -(2x+3)" for "x<-3/2):}" [2]"$

Use the distributive property to eliminate the ()s:

$f(x) = {(3 - 2x-3" for "x>=-3/2),(3 +2x+3" for "x<-3/2):}" [3]"$

Equation [3] is the desired piece-wise function.

How do you write f(x) = 3 - |2x + 3|f(x)=3−|2x+3|f(x) = 3 - |2x + 3| as a piecewise function?