Recall that sec x = . This means that sec x is undefined at any value where cos x = 0.
Also recall that cos = 0 and cos = 0. Therefore, sec x is continuous everywhere on
the interval because the interval does not include the endpoints.
- The function is continuous for k = .
In order for a function f(x) to be continuous at a point x = c, it must fulfill all three of the
following conditions:
Condition 1: f(c) exists.
Condition 2: f(x) exists.
Condition 3: f(x) = f(c)
We will need to find a value, or values, of k that enables f(x) to satisfy each condition.
Condition 1: f(4) = 0
Condition 2: f(x) = 0 and f(x) = 16k − 9. In order for the limit to exist, the two limits
must be the same. If we solve 16k − 9 = 0, we get k = .
Condition 3: If we now let k = , f(x) = 0 = f(4). Therefore, the solution is k = .
- The removable discontinuity is at .
A removable discontinuity occurs when you have a rational expression with common factors
in the numerator and denominator. Because these factors can be cancelled, the discontinuity is
“removable.” In practical terms, this means that the discontinuity occurs where there is a
“hole” in the graph. If we factor f(x) = , we get f(x) = . If we cancel
the common factor, we get f(x) = . Now, if we plug in x = 3, we get f(x) = . Therefore,
