How do you draw the graph of a modulus function where two modulus expressions are subtracted from each other?

Question

Let's say we have the expression $f(x) = \left|\left|x-1\right|\ -3\left|x+1\right|\right|$.

To draw it's graph, I would normally find the critical points ($x = 1, x = -1, x = \frac{-1}{2}$), then find the corresponding values of $y$ for each f(x), then join the points making the graph.

This would give us a graph looking like this. However, it turns out this graph is actually for $f(x) = \left|\left|x-1\right|\ +3\left|x+1\right|\right|$.

Normally, for the $3|x+1|$ expression, I would multiply the expression with the number outside it. But when that number is negative, the graph I was making was wrong, and my logic is incorrect.

The actual graph for $f(x) = \left|\left|x-1\right|\ -3\left|x+1\right|\right|$ is here.

So here is my question. How do you draw the graph of a modulus function where 2 modulus expressions are subtracted, and how do you go about multiplying a modulus expression (like $|x+1|$ with $-3$) in $f(x)$ with a negative number?

Answer 1

Draw the graph of $|x-1|-3|x+1|$ then reflect the parts of the graph which are below the $x$ axis with respect the $x$ axis. Draw the graph of the above function $f(x)=|x-1|-3|x+1|$ for the interval $(-\infty, -1]$, where $f(x)=-(x-1)+3(x+1)$. Then for the interval $[-1,1]$, and finally for $[1,+\infty)$.

This is the way.

Answer 2

For the graph of $f(x) = \left|\left|x-1\right|\ -3\left|x+1\right|\right|$, in the expression $-3|x+1|$you do NOT need to multiply $-3$ with $|x+1|$. Doing so will change the nature of the graph, giving you the wrong graph, as you can see here. This means that $-3|x+1| \neq |-3x-3|$. However, you can simplify the expression $-3|x+1|$ by writing it as $-|3x+3|$. This is correct, and does not change the graph, as you can see here, with both expressions giving the same graph.

To draw the graph of $f(x) = \left|\left|x-1\right|\ -3\left|x+1\right|\right|$, your conventional methods are correct. Just don't make the mistake of simplifying the expression by trying to bring $-3$ inside the the expression $|x+1|$.

User kmitov is correct. you have to draw the graph of $y = f(x)$ first, then draw the graph of $y = |f(x)|$

For $y = |x-1| - 3|x+1|$, the critical points are $-1$ and $1$. Find $y$ at each $f(x)$. Here, for $f(-1) =2$ and $f(1) = -6$. Plot the points. Your graph should look like this. Then find $f(x)$ at any $x < -1$ and $x > 1$ and join them to the graph, extending them till $-\infty$ (we know the nature of the graph does not change at points other than critical points). Your graph should look like this. This is the graph of $y = |x-1| - 3|x+1|$.

Now, we just have the draw the graph of $y =|f(x)|$. To do this, simply reflect the parts of the graph of $y = |x-1| - 3|x+1|$ below the $x$ axis (in the $-y$ axis) into the $+y$ axis. To check the abscissae where you reflect the graph (other than critical points), just equate $f(x) = 0$ and find the x coordinates (here, they are $-2$ and $\frac{-1}{2}$. This finally gives you the graph of the original expression $f(x) = \left|\left|x-1\right|\ -3\left|x+1\right|\right|$. The graph should look like this