Full solution

Q. Determine whether the function

f(x)

is continuous at

x=6

\newline

f(x)=\left\{\begin{array}{ll} 16-x^{2}, & x<6 \\ -2-3 x, & x \geq 6 \end{array}\right.

\newline

f(x)

is discontinuous at

x=6

\newline

f(x)

is continuous at

x=6

Check Function Definition: To determine if the function $f(x)$ is continuous at $x=6$ , we need to check three conditions: $\newline$ $1$ . The function is defined at $x=6$ . $\newline$ $2$ . The limit of $f(x)$ as $x$ approaches $6$ exists. $\newline$ $3$ . The limit of $f(x)$ as $x$ approaches $6$ is equal to the function value at $x=6$ . $\newline$ Let's start by checking if the function is defined at $x=6$ .
Evaluate $f(6)$ : The function $f(x)$ is defined piecewise, with one expression for $x < 6$ and another for $x \geq 6$ . At $x=6$ , the function is defined by the second piece, $f(x) = -2 - 3x$ . Let's evaluate $f(6)$ . $\newline$ $f(6) = -2 - 3(6) = -2 - 18 = -20$ . $\newline$ So, the function is defined at $x=6$ and $f(6) = -20$ .
Calculate Left Limit: Next, we need to find the limit of $f(x)$ as $x$ approaches $6$ from the left, which is denoted as $\lim_{x\to6^-} f(x)$ . For $x < 6$ , $f(x) = 16 - x^2$ . Let's calculate the limit. $\newline$ $\lim_{x\to6^-} f(x) = \lim_{x\to6^-} (16 - x^2) = 16 - (6)^2 = 16 - 36 = -20$ .
Calculate Right Limit: Now, we need to find the limit of $f(x)$ as $x$ approaches $6$ from the right, which is denoted as $\lim_{x\to6^+} f(x)$ . For $x \geq 6$ , $f(x) = -2 - 3x$ . Let's calculate the limit. $\newline$ $\lim_{x\to6^+} f(x) = \lim_{x\to6^+} (-2 - 3x) = -2 - 3(6) = -2 - 18 = -20$ .
Check Limit Existence: 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 $6$ exists and is equal to $-20$ .
Compare Limit and Function Value: Finally, we compare the limit of $f(x)$ as $x$ approaches $6$ to the function value at $x=6$ . Since both the limit and the function value at $x=6$ are equal to $-20$ , the function $f(x)$ is continuous at $x=6$ .