Determine whether the function f(x) is continuous at x=2. f(x)={[1-3x^(2)",",x 2],[-5-3x",",x

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Determine whether the function  f(x) is continuous at  x=2.  f(x)={[1-3x^(2)",",x > 2],[-5-3x",",x <= 2]:}  f(x) is discontinuous at  x=2  f(x) is continuous at  x=2

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Check Function Definition: To determine if the function f(x) is continuous at x=2, we need to check if the following three conditions are met: 1. The function is defined at x=2. 2. The limit of f(x) as x approaches 2 exists. 3. The limit of f(x) as x approaches 2 is equal to the function value at x=2.Find Limit Left: First, let's check if the function is defined at x=2. We have two expressions for f(x), one for x > 2 and one for x <= 2. Since we are looking at x=2, we will use the expression for x <= 2. f(2) = -5 - 3(2) = -5 - 6 = -11. The function is defined at x=2.Find Limit Right: Next, we need to find the limit of f(x) as x approaches 2 from the left (x -> 2-). For x <= 2, f(x) = -5 - 3x. lim (x -> 2-) f(x) = lim (x -> 2-) (-5 - 3x) = -5 - 3(2) = -11.Compare Limits: Now, we need to find the limit of f(x) as x approaches 2 from the right (x -> 2+). For x > 2, f(x) = 1 - 3x^2. lim (x -> 2+) f(x) = lim (x -> 2+) (1 - 3x^2) = 1 - 3(2)^2 = 1 - 3(4) = 1 - 12 = -11.Verify Continuity: Since the limit from the left and the limit from the right both exist and are equal to each other, the limit of f(x) as x approaches 2 exists and is equal to -11.Finally, we compare the limit of f(x) as x approaches 2, which is -11, to the function value at x=2, which is also -11. Since these two values are equal, the function is continuous at x=2.

Determine whether the function $f(x)$ is continuous at $x=2$ . $\newline$ $f(x)=\left\{\begin{array}{ll} 1-3 x^{2}, & x>2 \\ -5-3 x, & x \leq 2 \end{array}\right.$ $\newline$ $f(x)$ is discontinuous at $x=2$ $\newline$ $f(x)$ is continuous at $x=2$

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