How to define $y= |x+2|+|x-3|$ in a piecewise manner

Question

I need to define the function $y= |x+2|+|x-3|$ over the relevant intervals, but I am not entirely sure what this entails. How do I find the needed intervals? Plugging in different values gives me an idea of the shape of the graph, but I don't know how to use that to actually determine the piecewise definition of $y$.

Any guidance would be appreciated.

Answer 1

Firstly, apologies for the somewhat bad sketch.

If you sketch $y=|x+2|$ and $y=|x-3|$ on the same set of axes, notice the three intervals where the function $y=|x+2|+|x-3|$ will be defined differently. The dotted arrows in the sketch above should help with that.

Can you see where to go from here?

Edit: Your function (and what I'm fairly sure the question is asking for) may look something like this when you're done: $$y=\begin{cases}\text{something}, & x\leq-2\\ \text{something else}, & -2<x<3\\ \text{something else again}, & x\geq3\end{cases}$$

Answer 2

But I am not entirely sure what this entails.

It entails investigating the nature of $y=|x+2|+|x-3|$ by using your knowledge of $|x|$ which is defined as follows: $$ |x|= \begin{cases} x &\text{if $x \geq 0$},\\ -x &\text{if $x<0$}. \end{cases} $$ Similarly, for $|x+2|$, we have $$ |x+2|= \begin{cases} x+2 &\text{if $x \geq -2$},\\ -(x+2) &\text{if $x<-2$}. \end{cases}\tag{1} $$ Finally, for $|x-3|$, we have $$ |x-3|= \begin{cases} x-3 &\text{if $x \geq 3$},\\ -(x+3) &\text{if $x<3$}. \end{cases}\tag{2} $$ What intervals do we need to consider to graph $y$? A little bit of investigation reveals that we largely need to consider three intervals:

• Interval 1: $\quad x<-2$
• Interval 2: $\quad -2\leq x<3$
• Interval 3: $\quad x\geq 3$

Now use $(1)$ and $(2)$ to determine what $y$ will look like on these intervals:

For $x<-2$: $$ y=|x+2|+|x-3|=-(x+2)-(x-3)=-x-2-x+3=-2x+1. $$ For $-2\leq x<3$: $$ y=|x+2|+|x-3|=(x+2)-(x-3)=5. $$ For $x\geq 3$: $$ y=|x+2|+|x-3|=x+2+x-3=2x-1. $$ Hence, we have the following: $$ y=|x+2|+|x-3|= \begin{cases} -2x+1 &\text{if $x<-2$},\\ 5 &\text{if $-2\leq x<3$},\\ 2x-1 &\text{if $x\geq 3$}. \end{cases} $$ Your graph will look like this:

Answer 3

you can use the geometric interpretation of the absolute value function $|a-b|$ as the distance between the points represented $a$ and $b$ on the number line. that is rewrite $y = |x+2| + |x-3|$ as $y=|x-(-2)| + |x-3|.$ now interpret $y$ as the sum of the distances between the point $x$ to the points $-2$ and $3.$

with that, it becomes clear now that if $x$ is between $-2$ and $3,$ then $y$ is the constant $5,$ the distance between the points $-2$ and $3.$ if $x > 3,$ then $y$ is twice the distance from the midpoint $\frac 12,$ that is $2(x - \frac12) = 2x - 1.$ similarly if $x < -2, y = 1- 2x.$ in fact the graph of $y$ is symmetric about $x = \frac 12.$

so finally we have $$|x+2| +|x - 3| = \begin{cases} 1 - 2x & \text{ if }x \le -2\\5 & \text{ if } -2 < x < 3\\2x-1 & \text{ if } 3 \le x. \end{cases}$$