when doing absolute value functions (modulus), how do you know what domain to look at ?
e.g. y = -|3x + 1|,
and also the addition and subtraction ones? y = |x-3| - |x+2|
Remember that for any modulus function:
|=\left\{\begin{array}{lr}f(x)&f(x)\geq 0\\-f(x)&f(x)<0\end{array}\right.)
So, for your first one, we need to find out when the function is less than and greater than zero:
So, the function can be broken into a two piece hybrid:
&x\geq\frac{-1}{3}\\ -(-(3x+1))&x<\frac{-1}{3}\end{array}\right.<br />\implies y=\left\{\begin{array}{lr}-3x-1&x\geq\frac{-1}{3}\\ 3x+1&x<\frac{-1}{3}\end{array}\right.)
For your other one, let's consider the two parts separately as above.
For the first one (|x-3|), we have that x-3 when x>=3 and 3-x when x<3 (using the above working), and for |x+2| we have x+2 when x>=-2 and -x-2 when x<-2.
Now, when we combine them, we need to consider when these domains intersect. So, we have three cases:
1) x>=3, in which case both modulus are positive
2) -2<=x<3, in which case the first modulus is negative, but the second one is positive still.
3) x<-2, in which case both the modulus are negative.
Using this, we can construct a hybrid function for the three pieces:
&x\geq 3\\-(x-3)-(x+2)&-2\leq x< 3\\-(x-3)-(-(x+2))&x<-2 \end{array}\right.=\left\{\begin{array}{lr}-5&x\geq 3\\1-2x&-2\leq x< 3\\5&x<-2 \end{array}\right.)
(NOTE: I'm pretty sure general convention is that as you go from left to right in your domain, you go top to bottom in the piecewise function. I didn't do this, and I'm too lazy to fix that, but thoughts for next time.
)
You can also extend this to n additions of modulus functions, like so:
y=|x-5|+|x+1|-|x-2|
Using the theory above, see if you can figure this question out. I'll type up a solution in spoiler tags in a second.
Solution to above - try it first!!
Okay, so once again we consider the above three separately (I have run through it much quicker than above):
Now, we find the intersection points. At a glance we have three critical points (-1, 2, 5), so we have 4 intersection intervals:
x<-1, in which case all are negative
-1<=x<2, in which case |x+1| is positive and the other two are negative
2<=x<5, in which case |x-5| is negative and the other two are positive
x>=5, in which case all are positive.
So, this meaty thing gives us: (remember, y=|x-5|+|x+1|-|x-2|)
+(-(x+1))-(-(x-2)),&x<-1\\-(x-5)+(x+1)-(-(x-2)),&-1\leq x<2\\-(x-5)+(x+1)-(x-2)&,2\leq x<5\\(x-5)+(x+1)-(x-2),&x\geq 5\end{array}\right.<br />\\ =\left\{\begin{array}{lr}-x+5-x-1+x-2,&x<-1\\-x+5+x+1+x-2,&-1\leq x<2\\-x+5+x+1-x+2,&2\leq x<5\\x-5+x+1-x+2,&x\geq 5\end{array}\right.<br />\\ =\left\{\begin{array}{lr}2-x,&x<-1\\x+4,&-1\leq x<2\\8-x,&2\leq x<5\\x-2,&x\geq 5\end{array}\right.)
Looking at wolfram, this hybrid function looks very similar to the graph they've provided, and so we can say with good confidence that we're right.
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