Questions: Suppose f(x) is defined as shown below. a. Use the continuity checklist to show that f is not continuous at 1. b. Is f continuous from the left or right at 1? c. State the interval(s) of continuity. f(x) = x^2 + 4x if x >= 1 4x if x < 1 a. Why is f not continuous at 1? A. f(1) is not defined. B. The limit of f(x) as x approaches 1 does not exist. C. Although the limit of f(x) as x approaches 1 exists, it does not equal f(1).

Suppose f(x) is defined as shown below.
a. Use the continuity checklist to show that f is not continuous at 1.
b. Is f continuous from the left or right at 1?
c. State the interval(s) of continuity.
f(x) =
x^2 + 4x if x >= 1
4x if x < 1

a. Why is f not continuous at 1?
A. f(1) is not defined.
B. The limit of f(x) as x approaches 1 does not exist.
C. Although the limit of f(x) as x approaches 1 exists, it does not equal f(1).

Transcript text: Suppose $f(x)$ is defined as shown below. a. Use the continuity checklist to show that $f$ is not continuous at 1 . b. Is f continuous from the left or right at 1 ? c. State the interval(s) of continuity. \[ f(x)=\left\{\begin{array}{ll} x^{2}+4 x & \text { if } x \geq 1 \\ 4 x & \text { if } x<1 \end{array}\right. \] a. Why is f not continuous at 1 ? A. $f(1)$ is not defined. B. $\lim _{x \rightarrow 1} f(x)$ does not exist C. Although $\lim _{x \rightarrow 1} f(x)$ exists, it does not equal $f(1)$.

Solution

Solution Steps

Step 1: Check the definition of $ f(1) $

To determine if $ f $ is continuous at $ x = 1 $, first evaluate $ f(1) $. Since $ x = 1 $ falls under the condition $ x \geq 1 $, we use the first piece of the function: \[ f(1) = (1)^2 + 4(1) = 1 + 4 = 5. \] Thus, $ f(1) $ is defined and equals 5.

Step 2: Compute the left-hand limit $ \lim_{x \to 1^-} f(x) $

For $ x < 1 $, the function is defined as $ f(x) = 4x $. Therefore, the left-hand limit is: \[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} 4x = 4(1) = 4. \]

Step 3: Compute the right-hand limit $ \lim_{x \to 1^+} f(x) $

For $ x \geq 1 $, the function is defined as $ f(x) = x^2 + 4x $. Therefore, the right-hand limit is: \[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (x^2 + 4x) = (1)^2 + 4(1) = 1 + 4 = 5. \]

Step 4: Compare the left-hand limit, right-hand limit, and $ f(1) $

The left-hand limit is $ 4 $, the right-hand limit is $ 5 $, and $ f(1) = 5 $. Since the left-hand limit and right-hand limit are not equal ($ 4 \neq 5 $), the limit $ \lim_{x \to 1} f(x) $ does not exist. Therefore, $ f $ is not continuous at $ x = 1 $.

Step 5: Determine continuity from the left or right at $ x = 1 $

$ f $ is continuous from the right at $ x = 1 $ because $ \lim_{x \to 1^+} f(x) = f(1) = 5 $.
$ f $ is not continuous from the left at $ x = 1 $ because $ \lim_{x \to 1^-} f(x) = 4 \neq f(1) = 5 $.

Step 6: State the interval(s) of continuity

The function $ f(x) $ is continuous for all $ x \neq 1 $. Specifically:

For $ x < 1 $, $ f(x) = 4x $ is a linear function and is continuous everywhere.
For $ x > 1 $, $ f(x) = x^2 + 4x $ is a polynomial and is continuous everywhere.

Thus, the intervals of continuity are $ (-\infty, 1) $ and $ (1, \infty) $.