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

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Determine whether the function  f(x) is continuous at  x=5.  f(x)={[3+x^(2)",",x <= 5],[18+2x",",x > 5]:}  f(x) is continuous at  x=5  f(x) is discontinuous at  x=5

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Check Function Definition: To determine if the function f(x) is continuous at x=5, we need to check three conditions: 1. The function is defined at x=5. 2. The limit of f(x) as x approaches 5 exists. 3. The limit of f(x) as x approaches 5 is equal to the function value at x=5.Find Left Limit: First, let's check if the function is defined at x=5. We have two expressions for f(x), one for x <= 5 and one for x > 5. Since 5 is included in the domain of the first expression, we can use it to find the value of f(5). f(5) = 3 + 5^2 f(5) = 3 + 25 f(5) = 28 The function is defined at x=5.Find Right Limit: Next, we need to find the limit of f(x) as x approaches 5 from the left (x -> 5^-). We use the expression for x <= 5. lim (x -> 5^-) f(x) = lim (x -> 5^-) (3 + x^2) lim (x -> 5^-) f(x) = 3 + (5^-)^2 lim (x -> 5^-) f(x) = 3 + 25 lim (x -> 5^-) f(x) = 28Compare Limits: Now, we need to find the limit of f(x) as x approaches 5 from the right (x -> 5^+). We use the expression for x > 5. lim (x -> 5^+) f(x) = lim (x -> 5^+) (18 + 2x) lim (x -> 5^+) f(x) = 18 + 2(5^+) lim (x -> 5^+) f(x) = 18 + 10 lim (x -> 5^+) f(x) = 28Verify Continuity: Since the limit from the left and the limit from the right both exist and are equal, the limit of f(x) as x approaches 5 exists and is equal to 28.Finally, we compare the limit of f(x) as x approaches 5 to the function value at x=5. Since both are equal to 28, the function is continuous at x=5.

Determine whether the function $f(x)$ is continuous at $x=5$ . $\newline$ $f(x)=\left\{\begin{array}{ll} 3+x^{2}, & x \leq 5 \\ 18+2 x, & x>5 \end{array}\right.$ $\newline$ $f(x)$ is continuous at $x=5$ $\newline$ $f(x)$ is discontinuous at $x=5$

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