Determine whether the function f(x) is continuous at x=-3. f(x)={[4+2x^(2)",",x -3]:} f(x) is continuous at x=-3 f(x) is discontinuous at x=-3

Check Conditions: To determine if the function f(x) is continuous at x=-3, we need to check if the following three conditions are met: 1. f(-3) is defined. 2. The limit of f(x) as x approaches -3 exists. 3. The limit of f(x) as x approaches -3 is equal to f(-3). First, we will find the value of f(-3) using the appropriate piece of the function.Find f(-3): Since x = -3 falls in the interval where x -3^-). Since the function is the same for x -3^+). For x > -3, we use the second piece of the function: f(x) = 13 - 3x. Substitute x = -3 into the second piece of the function: limit as x -> -3^+ of f(x) = 13 - 3(-3). Calculate the right-hand limit: limit as x -> -3^+ of f(x) = 13 + 9 = 22.Verify Continuity: Since the left-hand limit and the right-hand limit as x approaches -3 are both equal to 22, and f(-3) is also 22, all three conditions for continuity are satisfied. Therefore, the function f(x) is continuous at x = -3.

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$Determine whether the function f(x) is continuous at x=-3. f(x)={[4+2x^(2)",",x <= -3],[13-3x",",x > -3]:} f(x) is continuous at x=-3 f(x) is discontinuous at x=-3$

Determine whether the function $f(x)$ is continuous at $x=-3$ . $\newline$ $f(x)=\left\{\begin{array}{ll} 4+2 x^{2}, & x \leq-3 \\ 13-3 x, & x>-3 \end{array}\right.$ $\newline$ $f(x)$ is continuous at $x=-3$ $\newline$ $f(x)$ is discontinuous at $x=-3$

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Intermediate Value Theorem

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Q. Determine whether the function

f(x)

is continuous at

x=-3

\newline

f(x)=\left\{\begin{array}{ll} 4+2 x^{2}, & x \leq-3 \\ 13-3 x, & x>-3 \end{array}\right.

\newline

f(x)

is continuous at

x=-3

\newline

f(x)

is discontinuous at

x=-3

Check Conditions: To determine if the function $f(x)$ is continuous at $x=-3$ , we need to check if the following three conditions are met: $\newline$ $1$ . $f(-3)$ is defined. $\newline$ $2$ . The limit of $f(x)$ as $x$ approaches $-3$ exists. $\newline$ $3$ . The limit of $f(x)$ as $x$ approaches $-3$ is equal to $f(-3)$ . $\newline$ First, we will find the value of $f(-3)$ using the appropriate piece of the function.
Find $f(-3)$ : Since $x = -3$ falls in the interval where $x \leq -3$ , we use the first piece of the function to find $f(-3)$ . Substitute $x = -3$ into the first piece of the function: $f(-3) = 4 + 2(-3)^2$ . Calculate $f(-3)$ : $f(-3) = 4 + 2(9) = 4 + 18 = 22$ .
Left-hand Limit: Next, we need to find the limit of $f(x)$ as $x$ approaches $-3$ from the left side ( $x \to -3^-$ ). $\newline$ Since the function is the same for $x \leq -3$ , the left-hand limit is the same as $f(-3)$ . $\newline$ Therefore, the left-hand limit as $x$ approaches $-3$ is $22$ .
Right-hand Limit: Now, we need to find the limit of $f(x)$ as $x$ approaches $-3$ from the right side ( $x \to -3^+$ ). $\newline$ For x > -3, we use the second piece of the function: $f(x) = 13 - 3x$ . $\newline$ Substitute $x = -3$ into the second piece of the function: limit as $x \to -3^+$ of $f(x) = 13 - 3(-3)$ . $\newline$ Calculate the right-hand limit: limit as $x \to -3^+$ of $x$ $0$ .
Verify Continuity: Since the left-hand limit and the right-hand limit as $x$ approaches $-3$ are both equal to $22$ , and $f(-3)$ is also $22$ , all three conditions for continuity are satisfied. $\newline$ Therefore, the function $f(x)$ is continuous at $x = -3$ .