Determine whether the function f(x) is continuous at x=-3. f(x)={[18-x^(2)",",x -3]:} f(x) is discontinuous at x=-3 f(x) is continuous at x=-3

Check Function Definition: To determine if the function f(x) is continuous at x=-3, we need to check three conditions: 1. The function is defined at x=-3. 2. The limit of f(x) as x approaches -3 exists. 3. The limit of f(x) as x approaches -3 is equal to the function value at x=-3.Find Left Limit: First, let's check if the function is defined at x=-3. We look at the piece of the function that applies when x is less than or equal to -3, which is f(x) = 18 - x^2. Plugging in x=-3, we get f(-3) = 18 - (-3)^2 = 18 - 9 = 9. So, the function is defined at x=-3.Find Right Limit: Next, we need to find the limit of f(x) as x approaches -3 from the left side (from values smaller than -3). Since the function for x -3 is f(x) = 15 + 3x. Plugging in x=-3, we get the limit as x approaches -3 from the right to be 15 + 3(-3) = 15 - 9 = 6.Check Continuity: Since the limit from the left side as x approaches -3 is 9 and the limit from the right side as x approaches -3 is 6, the two limits are not equal. Therefore, the limit of f(x) as x approaches -3 does not exist.Because the limit of f(x) as x approaches -3 does not exist, the function f(x) is not continuous at x=-3. The function fails the second condition for continuity.

Resources

Testimonials

Plans

Resources

Testimonials

Plans

AI tutor

Welcome to Bytelearn!

Let’s check out your problem:

$Determine whether the function f(x) is continuous at x=-3. f(x)={[18-x^(2)",",x <= -3],[15+3x",",x > -3]:} f(x) is discontinuous at x=-3 f(x) is continuous at x=-3$

Determine whether the function $f(x)$ is continuous at $x=-3$ . $\newline$ $f(x)=\left\{\begin{array}{ll} 18-x^{2}, & x \leq-3 \\ 15+3 x, & x>-3 \end{array}\right.$ $\newline$ $f(x)$ is discontinuous at $x=-3$ $\newline$ $f(x)$ is continuous at $x=-3$

View step-by-step help

Home

Math Problems

Algebra 1

Dilations of functions

Full solution

Q. Determine whether the function

f(x)

is continuous at

x=-3

\newline

f(x)=\left\{\begin{array}{ll} 18-x^{2}, & x \leq-3 \\ 15+3 x, & x>-3 \end{array}\right.

\newline

f(x)

is discontinuous at

x=-3

\newline

f(x)

is continuous at

x=-3

Check Function Definition: To determine if the function $f(x)$ is continuous at $x=-3$ , we need to check three conditions: $\newline$ $1$ . The function is defined at $x=-3$ . $\newline$ $2$ . The limit of $f(x)$ as $x$ approaches $-3$ exists. $\newline$ $3$ . The limit of $f(x)$ as $x$ approaches $-3$ is equal to the function value at $x=-3$ .
Find Left Limit: First, let's check if the function is defined at $x=-3$ . We look at the piece of the function that applies when $x$ is less than or equal to $-3$ , which is $f(x) = 18 - x^2$ . Plugging in $x=-3$ , we get $f(-3) = 18 - (-3)^2 = 18 - 9 = 9$ . So, the function is defined at $x=-3$ .
Find Right Limit: Next, we need to find the limit of $f(x)$ as $x$ approaches $-3$ from the left side (from values smaller than $-3$ ). Since the function for $x \leq -3$ is $f(x) = 18 - x^2$ , we use this to find the limit. The limit as $x$ approaches $-3$ from the left is the same as the function value at $x=-3$ , which we already calculated as $9$ .
Compare Limits: Now, we need to find the limit of $f(x)$ as $x$ approaches $-3$ from the right side (from values greater than $-3$ ). The function for x > -3 is $f(x) = 15 + 3x$ . Plugging in $x=-3$ , we get the limit as $x$ approaches $-3$ from the right to be $15 + 3(-3) = 15 - 9 = 6$ .
Check Continuity: Since the limit from the left side as $x$ approaches $-3$ is $9$ and the limit from the right side as $x$ approaches $-3$ is $6$ , the two limits are not equal. Therefore, the limit of $f(x)$ as $x$ approaches $-3$ does not exist.
Check Continuity: Since the limit from the left side as $x$ approaches $-3$ is $9$ and the limit from the right side as $x$ approaches $-3$ is $6$ , the two limits are not equal. Therefore, the limit of $f(x)$ as $x$ approaches $-3$ does not exist.Because the limit of $f(x)$ as $x$ approaches $-3$ does not exist, the function $f(x)$ is not continuous at $-3$ $3$ . The function fails the second condition for continuity.