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I'm given $$ k(x)=\begin{cases} -5x+6,& x<4\\-4x+2,&x\ge 4\:\end{cases} $$ And I have to find $k^{-1}(x)$.

I found inverse of the both functions $$ -\frac{x-2}{4} $$ $$ -\frac{x-6}{5} $$

But what will their range be?

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    Plugging in $4$ in $-4x+2$ you get $-14$,now since the function is decreasing the range of the function is $(-\infty,-14]$ and the second is $(-14,\infty)$ – 2017-01-04
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    @kingW3 you have provided the range of the original function $k(x)$. Isn't the OP asking for the range of the inverse? – 2017-01-04

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Despite the change of its form at $k=4$, the function $k(x)$ is monotone decreasing on $\mathbb R$ and continuous at $x=4$. Hence $$k^{-1}(x)=\begin{cases}-\dfrac{x-2}{4}, & x\le -14 \\ -\dfrac{x-6}{5}, & x>-14\end{cases}$$ where $-14=k(4)$.

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So good job on finding your inverses:

$$ -\frac{x-2}{4} $$ $$ -\frac{x-6}{5} $$

Recall that the definition of a function maps $x$ values to $y$ values, or $k(x)$. Therefore, the inverse of a function $k(x)$ maps the $y$ values to the $x$ values.

So we must find our range of $x$ values:

Since $k(4) = -14$, $k^{-1}(-14) = 4$

So we have:

$$k^{-1}(x)=\begin{cases}-\dfrac{x-2}{4}, & x\le -14 \\ -\dfrac{x-6}{5}, & x>-14\end{cases}$$

Therefore, if we want the range of these inverse functions, we simply look at the domain of our original function $k(x)$.

Therefore, the range of the inverse functions are $(-\infty, 4)$ & $(4, \infty)$, respectively.