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Draw labeled graph of: $|x-1|-|3x-2|$. where $|\cdot|$ denoted Modulus or Absolute value function.

I tried braking graphs between intervals: $(-\infty,\frac{2}{3}),\ \big[\frac{2}{3},1),\ \big[1,\infty)$. Obtained three different graphs then combined them into one to obtain $|x-1|-|3x-2|$.

But that too cumbersome process, can this be done easier manner, than this?

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  • $\begingroup$ Not really. However, your intervals should be $(-\infty, 2/3)$, $[2/3, 1)$, and $[1, \infty)$, or some other union that includes the endpoints of the intervals. $\endgroup$ – N. F. Taussig Aug 18 '17 at 11:29
  • $\begingroup$ @N.F.Taussig yes, typo. $\endgroup$ – mathlover Aug 18 '17 at 11:32
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Such functions are piecewise linear, so that it suffices to evaluate the function at the "corners" and draw segments in between. A corner occurs where the argument of an absolute value goes to zero.

We have

$$3x-2=0\to f\left(\frac23\right)=\frac13,\\x-1=0\to f(1)=-1$$ and this is enough to draw in $\left[\dfrac23,1\right]$.

Outside this range, we have two half-lines and it is sufficient to know their slopes, $2$ on the left and $-2$ on the right. This is because the slope of $|x|$ is $-1$ then $1$ from left to right.

This is confirmed by a grapher:

enter image description here


Summary of the method:

For a linear combination of absolute values of linear binomials,

  • evaluate the sum at all the corner points, which occur at the zeroes of the binomials,

  • evaluate the left and right slopes, which are minus and plus the sum of the coefficients respectively,

  • draw the polyline.

For $$\sum a_k|b_kx+c_k|,$$

evaluate all

$$x_n=-\frac{b_n}{a_n}$$ then

$$y_n\sum a_k|b_kx_n+c_k|,$$ and link $(x_n,y_n)$, sorted on $x$.

The slopes of the extreme half-lines are $$\mp\sum a_k.$$

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