Question

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?

Answer 1

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

Explanation:

`f(x) = {(-6x-12,"when",x < -3),(k^2-5x,"when",x = -3),(6,"when",x > -3):}`

Function `f` is continuous at `x=-3` if and only if `lim_(xrarr-3)f(x)=f(-3)`.

Looking at the first branch of the function we see that

`lim_(xrarr-3^-)f(x) = lim_(xrarr-3)(-6x-12) = -6(-3)-12 = 18-12=6`.

Looking at the last branch, we see that

`lim_(xrarr-3^+)f(x) = lim_(xrarr-3)6 = 6`.

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

`lim_(xrarr-3) f(x) = 6`.

To make `f` continuous at `-3` we need to make `f(-3) = 6`.

Looking at the definition of `f` we see that `f(-3) = k^2-5(-3) = 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 = +-3i`.

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`.