PROBLEM 2. Is the function f(x) = continuous at x = 1?
Answer: Condition 1: f(1) = 9.
Condition 2: The f(x) = 9 and the f(x) = 9.
Therefore, the f(x) exists and is equal to 9.
Condition 3: f(x) = f (1) = 9.
The function satisfies all three conditions, so it is continuous at x = 1.
PROBLEM 3. For what value of a is the function f(x) = continuous at x = 4?
Answer: Because f(4) = 12, the function passes the first condition.
For Condition 2 to be satisfied, the f(x) = 4a + 5 must equal the f(x) = 12. So set 4a + 5 = 12. If
a = , the limit will exist at x = 4 and the other two conditions will also be fulfilled. Therefore, the value
a = makes the function continuous at x = 4.
PROBLEM 4. Where does the function f(x) = have: (a) an essential discontinuity; and (b)
a removable discontinuity?
Answer: If you factor the top and bottom of this fraction, you get
Thus, the function has an essential discontinuity at x = −4. If we then cancel the term (x − 5), and substitute
x = 5 into the reduced expression, we get f(5) = . Therefore, the function has a removable discontinuity
at
Note: Don’t confuse coordinate parentheses with interval
notation. In interval notation, square brackets include endpoints