Full solution

Q. Determine whether the function

f(x)

is continuous at

x=3

\newline

f(x)=\left\{\begin{array}{ll} 19-4 x^{2}, & x \geq 3 \\ -8-3 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 Left Limit: First, let's check if the function is defined at $x=3$ . We will use the piece of the function that applies when $x$ is greater than or equal to $3$ , which is $f(x) = 19 - 4x^2$ . $\newline$ Substitute $x = 3$ into the function to find $f(3)$ . $\newline$ $f(3) = 19 - 4(3)^2 = 19 - 4(9) = 19 - 36 = -17$ . $\newline$ The function is defined at $x=3$ , and $f(3) = -17$ .
Find Right Limit: Next, we need to find the limit of $f(x)$ as $x$ approaches $3$ from the left side ( $x < 3$ ). We will use the piece of the function that applies when $x$ is less than $3$ , which is $f(x) = -8 - 3x$ . $\newline$ Find the limit as $x$ approaches $3$ from the left: $\newline$ $\lim_{x \to 3^-} f(x) = -8 - 3(3) = -8 - 9 = -17$ .
Verify Continuity: Now, we need to find the limit of $f(x)$ as $x$ approaches $3$ from the right side ( $x \geq 3$ ). We will use the piece of the function that applies when $x$ is greater than or equal to $3$ , which is $f(x) = 19 - 4x^2$ . Find the limit as $x$ approaches $3$ from the right: $\lim_{x \to 3^+} f(x) = 19 - 4(3)^2 = 19 - 4(9) = 19 - 36 = -17$ .
Verify Continuity: Now, we need to find the limit of $f(x)$ as $x$ approaches $3$ from the right side ( $x \geq 3$ ). We will use the piece of the function that applies when $x$ is greater than or equal to $3$ , which is $f(x) = 19 - 4x^2$ . Find the limit as $x$ approaches $3$ from the right: $\lim_{x \to 3^+} f(x) = 19 - 4(3)^2 = 19 - 4(9) = 19 - 36 = -17$ . Since the limit from the left side as $x$ approaches $3$ is $x$ $2$ , and the limit from the right side as $x$ approaches $3$ is also $x$ $2$ , and the function value at $x$ $6$ is $x$ $2$ , all three conditions for continuity are met. Therefore, the function $f(x)$ is continuous at $x$ $6$ .