Solving Piecewise Functions with Two Absolute Values

[Loppyfoot]

Homework Statement

Express the function in piecewise form without using absolute values.

Homework Equations

g(x) = |x| + |x-1|

The Attempt at a Solution

Alright, so I can do this problem when there is only one absolute value piece, but there are two different ones; so I do not know where to begin. Are the breakpoints at 0 and 1? How do I go about generating a piece wise function when both of them are absolute values?

 

[Hootenanny]

Homework Statement

Express the function in piecewise form without using absolute values.

Homework Equations

g(x) = |x| + |x-1|

The Attempt at a Solution

Alright, so I can do this problem when there is only one absolute value piece, but there are two different ones; so I do not know where to begin. Are the breakpoints at 0 and 1? How do I go about generating a piece wise function when both of them are absolute values?

As you say, there are two 'breakpoints', therefore you will need to split the domain into three sections:

1. x<0
2. 0<x<1
3. x>1

 

[Loppyfoot]

Could you explain how you got those? I have always struggled with piecewise functions, and I want to master them. Also, how does one get the left side of the piecewise? (ie; not the inequality part)

 

[Hootenanny]

Could you explain how you got those? I have always struggled with piecewise functions, and I want to master them. Also, how does one get the left side of the piecewise? (ie; not the inequality part)

As you know, we only need to worry about the mod function when the argument becomes negative. So when x < 0, the arguments of both mod functions is negative. When 0 < x < 1, the argument of the first mod function is positive whilst the second is negative. Finally, when x > 1 the arguments of both mod functions are positive and we therefore don't need to worry about them any more.

 

[Loppyfoot]

The back of my good has this;
1-2x ; x<0
1; 0<x<1
2x-1; x>1
I understand that we do not need to worry about the mod functions after the one breakpoint, but how does my book come up with the "(1-2x, 1, and 2x-1)?

Thanks

 

[Hootenanny]

Let's consider the case of x < 0. In this case do you agree that the argument of both mod functions is negative?

How does the presence of the mod function affect the argument if it is negative?

 

[Loppyfoot]

Let's consider the case of x < 0. In this case do you agree that the argument of both mod functions is negative? So if x<0, wouldn't the absolute value be positive? Is this the mod function?

 

[Hootenanny]

Let's consider the case of x < 0. In this case do you agree that the argument of both mod functions is negative? So if x<0, wouldn't the absolute value be positive? Is this the mod function?

The mod function is defined as follows

\left|x\right| = \left\{\begin{array}{lll} x &amp; \text{for} &amp; x \geq 0 \\ -x &amp; \text{for} &amp; x &lt; 0 \end{array}\right.

Do you understand?

 

[Loppyfoot]

Ok, I see that now; that is the first mod function.

 

[VietDao29]

Ok, I see that now; that is the first mod function.

Well, I'm not sure if you really get the concept of absolute value of a (real) number. So I'm going over it again. If you find any steps, or explanation confusing, just don't hesitate to shout it out.

The absolute value of a (real) number a is defined as follow:

|a| = \left\{ \begin{array}{cl} a &amp; \mbox{, if } a \geq 0 \\ -a &amp; \mbox{, if } a &lt; 0 \end{array} \right.

Or:

|a| = \left\{ \begin{array}{cl} a &amp; \mbox{, if } a &gt; 0 \\ -a &amp; \mbox{, if } a \leq 0 \end{array} \right.

Since |0| = 0 = -0, so the 2 definitions are actually the same.

Since, the former one is more common, we'll be working with it. So, if a is non-negative, then just simply keep it, i.e: |a| = a. For example: |4| = 4, |0| = 0.

If a is negative, we then take the opposite (additive inverse) of it, i.e: |a| = -a. For example: |-7| = -(-7) = 7, |-100| = -(-100) = 100.

Or, roughly speaking, the absolute value of a number is always non-negative, and is the "distance" from that number to 0 (the origin), in the real number line.

--------------

Ok, I'll give you an example:

Example:

Express the following function in piecewise form.

f(x) = |x - a|, for a is any real constant.

The first step of this type of questions is to solve for x, where the function(s) inside the absolute signs are 0.

x - a = 0 => x = a.

So, the breakpoint is at a.

Split it into 2 cases:

1. x >= a:
   
   When x >= a, x - a is non-negative, right? So, we have: |x - a| = x - a​
2. x < a:
   
   When x < a, x - a is always negative, right? So, we have: |x - a| = -(x - a) = -x + a​

So, our piecewise function is:

|f(x)| = \left\{ \begin{array}{cl} x - a &amp; \mbox{, if } x \geq a \\ a - x &amp; \mbox{, if } x &lt; a \end{array} \right.

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Back to your question, after finding the 2 breakpoints, namely 0, and 1. We then, split it into 3 different parts:

1. x < 0:
   
   When x < 0, x is negative, right? So |x| = -x​
   When x < 0, x - 1 is also negative, right? So |x - 1| = -(x - 1) = -x + 1​
   So, when x < 0, your function will be (without absolute signs) f(x) = |x| + |x - 1| = -x + (-x + 1) = -2x + 1​
2. 0 <= x < 1: (pay attention to the <= sign, it should be less than or equal, not just less than, since you should consider x = 0 as well)
   
   This can be worked out in almost the same manner.​
3. x >= 1:
   
   This part should be simple too, just follow my example. :)​