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$Determine whether the function f(x) is continuous at x=2. f(x)={[2-x^(2)",",x < 2],[-8+3x",",x >= 2]:} f(x) is continuous at x=2 f(x) is discontinuous at x=2$

Determine whether the function $f(x)$ is continuous at $x=2$ . $\newline$ $f(x)=\left\{\begin{array}{ll} 2-x^{2}, & x<2 \\ -8+3 x, & x \geq 2 \end{array}\right.$ $\newline$ $f(x)$ is continuous at $x=2$ $\newline$ $f(x)$ is discontinuous at $x=2$

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Precalculus

Identify discrete and continuous random variables

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Q. Determine whether the function

f(x)

is continuous at

x=2

\newline

f(x)=\left\{\begin{array}{ll} 2-x^{2}, & x<2 \\ -8+3 x, & x \geq 2 \end{array}\right.

\newline

f(x)

is continuous at

x=2

\newline

f(x)

is discontinuous at

x=2

Check Function Definition: To determine if the function $f(x)$ is continuous at $x=2$ , we need to check three conditions: $\newline$ $1$ . The function is defined at $x=2$ . $\newline$ $2$ . The limit of $f(x)$ as $x$ approaches $2$ exists. $\newline$ $3$ . The limit of $f(x)$ as $x$ approaches $2$ is equal to the function value at $x=2$ .
Find Left Limit: First, let's check if the function is defined at $x=2$ . We look at the piece of the function that applies when $x$ is greater than or equal to $2$ , which is $f(x) = -8 + 3x$ . We substitute $x$ with $2$ to find $f(2)$ . $\newline$ $f(2) = -8 + 3(2) = -8 + 6 = -2$ . $\newline$ The function is defined at $x=2$ because $f(2) = -2$ .
Find Right Limit: Next, we need to find the limit of $f(x)$ as $x$ approaches $2$ from the left side ( $x < 2$ ). We use the piece of the function that applies when $x$ is less than $2$ , which is $f(x) = 2 - x^2$ . We substitute $x$ with a value that is very close to $2$ from the left, such as $1.999$ . $\newline$ $x$ $0$ $\newline$ However, we need to consider the exact value as $x$ approaches $2$ , not an approximation. $\newline$ $x$ $3$
Verify Continuity: Now, we need to find the limit of $f(x)$ as $x$ approaches $2$ from the right side ( $x \geq 2$ ). We use the piece of the function that applies when $x$ is greater than or equal to $2$ , which is $f(x) = -8 + 3x$ . We substitute $x$ with a value that is very close to $2$ from the right, such as $2.001$ . $x$ $0$ . Again, we need to consider the exact value as $x$ approaches $2$ , not an approximation. $x$ $3$ .
Verify Continuity: Now, we need to find the limit of $f(x)$ as $x$ approaches $2$ from the right side ( $x \geq 2$ ). We use the piece of the function that applies when $x$ is greater than or equal to $2$ , which is $f(x) = -8 + 3x$ . We substitute $x$ with a value that is very close to $2$ from the right, such as $2.001$ .
$x$ $0$ .
Again, we need to consider the exact value as $x$ approaches $2$ , not an approximation.
$x$ $3$ .Since the limit from the left side as $x$ approaches $2$ is $x$ $6$ and the limit from the right side as $x$ approaches $2$ is also $x$ $6$ , and both of these limits are equal to the function value at $2$ $0$ ( $2$ $1$ ), all three conditions for continuity are satisfied.
Therefore, the function $f(x)$ is continuous at $2$ $0$ .