Answer
$k^{-1}_1(x)=-x+3$ (Red dotted line graph in the image), which is inverse of the function $k_1(x)$ with domain $(-\infty \lt x \leq 3)$ (Red line graph in the image)
$k^{-1}_2(x)=x+3$ (Blue dotted line graph in the image), which is inverse of the function $k_2(x)$ with domain $(3 \leq x \lt \infty)$ (Blue line graph in the image)
Work Step by Step
We know that due to the definition of the inverse function, for a function to have inverse it should have the form that every $y$ value should get only one $x$ value, so we have to restrict either left-hand side(Red graph in the image) or right-hand side(Blue graph in the image).
So the domain of definition should be:
$(-\infty \lt x \leq 3)$ - Which is Red line graph in the image. Let's call this $h_1(x)$
or
$(3 \leq x \lt \infty)$ - Which is Blue line graph in the image. Let's call this $h_2(x)$
To calculate inverse of the function $6(x)=|x-3|$, we will follow the steps which is required to calculate it:
First write the function in terms of $y$ and $x$:
$y=|x-3|$
Then replace $y$ by $x$ and vice versa:
$x=|y-3|$
And at last, solve it for $y$:
$x=-(y-3)$ or $x=y-3$
$y_1=3-x$; $y_2=x+3$
So we have:
$k^{-1}_1(x)=-x+3$ (Red dotted line graph in the image), which is inverse of the function $k_1(x)$ with domain $(-\infty \lt x \leq 3)$ (Red line graph in the image)
$k^{-1}_2(x)=x+3$ (Blue dotted line graph in the image), which is inverse of the function $k_2(x)$ with domain $(3 \leq x \lt \infty)$ (Blue line graph in the image)
