Piecewise functions

Question

Hi, I am a little comfused by piecewise functions. Its going to be on my test and whats making me nervous is the absolute value. These are the ones that I need help creating the piecewise function: 

 

a) f(x)= |x|-2

b)f(x)=-|x+1|+2

c)f(x)= |x+3| -4

d) f(x)=|x+5|

Answer 1

Alex,

 

To understand piecewise functions, you must remember that a function is like a rule which tells you how to find something from something else. Some functions are simpler than others and can be stated by a single rule. Examples are x² or √x. Others are more complicated and require more than one rule depending on the value of x that is input. this is the basic idea behind piecewise functions. You give a rule, but it works only for certain values of x. Other values of x require a different rule. Examples of piecewise functions occur widely in real life. Utility billing or income tax rates, for instance. They are not just designed to make your study of algebra harder. they really are quite useful.

 

As for absolute value, there are several ways of thinking about it. For what you are doing, thinking of it as a piecewise function is best.

 

Remember that absolute value strips the sign off of a number and leaves it positive or zero. As I am sure you know,

 

|2| = 2, while

 

|−2| = 2, and

 

|0| = 0.

 

You can define absolute value as a piecewise function as follows:

 

|x| = x if x ≥ 0  (just ignore the sign as it is already positive or zero)

     = −x if x < 0  (change the sign if the number is negative to make it positive)

 

The second part is a little hard to see, but it is the way to write one function as a formula in terms of what you already have available to you.

 

Now, let's go back to your questions.

 

For a), the function is f(x) = |x| − 2. When you use what I just gave you, you can break this down into two parts or "pieces":

 

f(x) = x − 2 if x ≥ 0

 

      = −x − 2 if x < 0.

 

(I am omitting the big left brace ({) which comes after the equals sign because I do not have the formatting available to show this here.)

 

I will do one more, and hope you can finish the others on your own. If you need more help, you can always send another reply, but it may be answered by another tutor.

 

b) You are given f(x) = −|x + 1| + 2. To break the absolute value down, you can replace x by x + 1:

 

|x + 1| = x + 1 if x + 1 ≥ 0

           = −(x + 1) if x + 1 < 0.

 

After doing a little algebra, this can be rewritten as:

 

|x + 1| = x + 1 if x ≥ −1 and 

           = −x − 1 if x < −1.

 

The definition changes at x = −1, as you can see. The formula is different for x less than −1 than for larger x.

Going back to the original question, you can rewrite f(x) using what I just showed you as:

 

f(x) = −(x + 1) + 2  if x ≥ −1

 

      = (x + 1) + 2  if x < −1.  Notice that I used two minus signs make a positive sign on this part.

 

Now, simplify the two parts:

 

f(x) = −x + 1  if x ≥ −1

      = x + 3  if x < −1.

 

This will be the final answer simplified as much as possible.

 

Here are some hints for the others:

 

c) Same idea as in b), but there is no minus before the absolute value.

d) Even simpler than c) as less simplification will be required.

Answer 2

hey Alex, 

you really need not worry about it. 

whenever you have absolute value involved think about when the inside is positive and when it's negative. 

let's look at the first problem

f(x) =|x|-2

when x>0, then 

f(x) =x-2

when x=0, then 

f(x) =-2

when x<0, then 

f(x) =-x-2

now for the next question f(x) =-|x+1|+2

instead of saying when x>0 or x<0

we will discuss when x+1>0 or x+1<0

as I said we are always discussing whether or not the inside of the abs value if it is positive or if it is negative 

if it is positive just take out abs value bars without change. if negative multiply by negative so we always have positive quantity. absolute values always returns a positive output. 

so f(x) =-|x+1|+2

when x+1>0 (same thing as x>-1)

then f(x) =-(x+1)+2

if x+1=0 ( x=-1)  then f(x) =2

if x+1<0 ( same as x<-1) then f(x) = - (-(x+1))+2=x+1+2=x+3

PS: realize that I also discussed when the inside is 0 

same method with the rest 

good luck with the exma