What is the domain of k? $2|x-2|-3|x-3|=k$

Question

What is the domain of k?
$2|x-2|-3|x-3|=k$
Using the definition of absolute function, I obtain:
$|x-2|= \begin{cases} x-2, & x\ge2 \\ 2-x, & x<2\\ \end{cases}\\$
$|x-3|= \begin{cases} x-3, & x\ge3 \\ 3-x, & x<3 \\ \end{cases} \\$

For $x<2$
$x-5=k$

For $2\le x<3$
$5x-13=k$

For $x \ge 3$
$-x+5=k$

But I need to find the domain of k.

Answer 1

You need to say $$f(x<2)= x-5, f(2\le x<3)=5x-13, f(x\ge 3)= 5-x$$ It is a continuous function such that $f(\pm \infty)=-\infty$ its plot suggests that and $f_{max}=f(3)=2$, see below. So the domain of $k$, is the range of $f(x)$ and it is $(-\infty, 2]$.

Answer 2

From the $3$ rules that you worked out, the graph of $2|x-2|-3|x-3|$ increases on $(-\infty, 3]$ and then decreases from $[3, \infty)$.

Hence the possible values of $k$ are $(-\infty, f(3)) = (-\infty, 2]$.

Answer 3

• If $ -\infty <x≤2$, then you have

$$k=2(2-x)-3(3-x)=x-5 \Longrightarrow - \infty <k=x-5≤-3$$

• If $2<x≤3$, then you have

$$k=2(x-2)-3(3-x)=5x-13 \Longrightarrow -3<k=5x-13≤2$$

• If $3<x<\infty$, then you have

$$k=2(x-2)-3(x-3)=-x+5 \Longrightarrow 2>k=-x+5> - \infty \Longrightarrow -\infty <k<2$$

Finally, you get

$$k \in (- \infty, -3] ∪ (-3, 2] ∪ (-\infty, 2)= (-\infty,2].$$