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?

1 Answer
Oct 20, 2016

Answer:

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#.