# For what value(s) of k is the function f(x) continuous at x = -3 given f(x) = -6x - 12 when x < -3, f(x) = k^2 - 5k when x = -3 and f(x) = 6 when x > -3?

Oct 20, 2016

There is no real value of $k$ that will make $f$ continuous at $- 3$.. If we are allowed imaginary solutions, then we need $k = \pm 3 i$.

#### Explanation:

$f \left(x\right) = \left\{\begin{matrix}- 6 x - 12 & \text{when" & x < -3 \\ k^2-5x & "when" & x = -3 \\ 6 & "when} & x > - 3\end{matrix}\right.$

Function $f$ is continuous at $x = - 3$ if and only if ${\lim}_{x \rightarrow - 3} f \left(x\right) = f \left(- 3\right)$.

Looking at the first branch of the function we see that

${\lim}_{x \rightarrow - {3}^{-}} f \left(x\right) = {\lim}_{x \rightarrow - 3} \left(- 6 x - 12\right) = - 6 \left(- 3\right) - 12 = 18 - 12 = 6$.

Looking at the last branch, we see that

${\lim}_{x \rightarrow - {3}^{+}} f \left(x\right) = {\lim}_{x \rightarrow - 3} 6 = 6$.

Because the left and right limits are both $6$, we have

${\lim}_{x \rightarrow - 3} f \left(x\right) = 6$.

To make $f$ continuous at $- 3$ we need to make $f \left(- 3\right) = 6$.

Looking at the definition of $f$ we see that $f \left(- 3\right) = {k}^{2} - 5 \left(- 3\right) = {k}^{2} + 15$

In order to make the function continuous at $- 3$, we need to make

${k}^{2} + 15 = 6$, so

${k}^{2} = - 9$ and $k$ is an imaginary number.

If we are allowed imaginary solutions, we need $k = \pm 3 i$.

If we are restricted to real number solutions, then the proper answer is: there is no real value of $k$ that will make $f$ continuous at $- 3$.