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$Determine whether the function f(x) is continuous at x=3. f(x)={[7+x^(2)",",x >= -3],[12-2x",",x < -3]:} f(x) is discontinuous at x=3 f(x) is continuous at x=3$

Determine whether the function $f(x)$ is continuous at $x=3$ . $\newline$ $f(x)=\left\{\begin{array}{ll} 7+x^{2}, & x \geq-3 \\ 12-2 x, & x<-3 \end{array}\right.$ $\newline$ $f(x)$ is discontinuous at $x=3$ $\newline$ $f(x)$ is continuous at $x=3$

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Q. Determine whether the function

f(x)

is continuous at

x=3

\newline

f(x)=\left\{\begin{array}{ll} 7+x^{2}, & x \geq-3 \\ 12-2 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 if the following three conditions are met: $\newline$ $1$ . 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 Limit Approaching $3$ : First, let's check if the function is defined at $x=3$ . Since $x=3$ is greater than or equal to $-3$ , we use the first piece of the function, which is $7 + x^2$ . Plugging in $x=3$ , we get $f(3) = 7 + 3^2 = 7 + 9 = 16$ . So, the function is defined at $x=3$ .
Compare Limit and Function Value: Next, we need to find the limit of $f(x)$ as $x$ approaches $3$ from the left and from the right. For $x$ approaching $3$ from the right, we use the first piece of the function, $7 + x^2$ . The limit as $x$ approaches $3$ from the right is $\lim_{x\to3^+} 7 + x^2 = 7 + 3^2 = 16$ .
Compare Limit and Function Value: Next, we need to find the limit of $f(x)$ as $x$ approaches $3$ from the left and from the right. For $x$ approaching $3$ from the right, we use the first piece of the function, $7 + x^2$ . The limit as $x$ approaches $3$ from the right is $\lim_{x\to3+} 7 + x^2 = 7 + 3^2 = 16$ .For $x$ approaching $3$ from the left, we still use the first piece of the function, $7 + x^2$ , because the second piece is for $x$ $2$ , which does not include values around $x$ $3$ . The limit as $x$ approaches $3$ from the left is $x$ $6$ .
Compare Limit and Function Value: Next, we need to find the limit of $f(x)$ as $x$ approaches $3$ from the left and from the right. For $x$ approaching $3$ from the right, we use the first piece of the function, $7 + x^2$ . The limit as $x$ approaches $3$ from the right is $\lim_{x\to3^+} 7 + x^2 = 7 + 3^2 = 16$ .For $x$ approaching $3$ from the left, we still use the first piece of the function, $7 + x^2$ , because the second piece is for $x$ $2$ , which does not include values around $x$ $3$ . The limit as $x$ approaches $3$ from the left is $x$ $6$ .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 $3$ exists and is equal to $3$ $0$ .
Compare Limit and Function Value: Next, we need to find the limit of $f(x)$ as $x$ approaches $3$ from the left and from the right. For $x$ approaching $3$ from the right, we use the first piece of the function, $7 + x^2$ . The limit as $x$ approaches $3$ from the right is $\lim_{x\to3^+} 7 + x^2 = 7 + 3^2 = 16$ .For $x$ approaching $3$ from the left, we still use the first piece of the function, $7 + x^2$ , because the second piece is for $x$ $2$ , which does not include values around $x$ $3$ . The limit as $x$ approaches $3$ from the left is $x$ $6$ .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 $3$ exists and is equal to $3$ $0$ .Finally, we compare the limit of $f(x)$ as $x$ approaches $3$ , which is $3$ $0$ , to the function value at $x$ $3$ , which we found to be $3$ $0$ . Since these two values are equal, the function is continuous at $x$ $3$ .