Determine whether the function f(x) is continuous at x=3. f(x)={[4-2x^(2)",",x 3],[-11-2x",",x

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Determine whether the function  f(x) is continuous at  x=3.  f(x)={[4-2x^(2)",",x > 3],[-11-2x",",x <= 3]:}  f(x) is discontinuous at  x=3  f(x) is continuous at  x=3

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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 have two expressions for f(x), one for x > 3 and one for x <= 3. Since we are looking at x=3, we will use the expression for x <= 3. f(3) = -11 - 2(3) = -11 - 6 = -17. 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 (x -> 3^-). For x <= 3, f(x) = -11 - 2x. lim (x -> 3^-) f(x) = lim (x -> 3^-) (-11 - 2x) = -11 - 2(3) = -17.Determine Continuity: Now, we need to find the limit of f(x) as x approaches 3 from the right (x -> 3^+). For x > 3, f(x) = 4 - 2x^2. lim (x -> 3^+) f(x) = lim (x -> 3^+) (4 - 2x^2) = 4 - 2(3)^2 = 4 - 18 = -14.We have found that the limit from the left is -17 and the limit from the right is -14. Since these two limits are not equal, the limit of f(x) as x approaches 3 does not exist. Therefore, the function f(x) is not continuous at x=3.

Determine whether the function $f(x)$ is continuous at $x=3$ . $\newline$ $f(x)=\left\{\begin{array}{ll} 4-2 x^{2}, & x>3 \\ -11-2 x, & x \leq 3 \end{array}\right.$ $\newline$ $f(x)$ is discontinuous at $x=3$ $\newline$ $f(x)$ is continuous at $x=3$

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