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$Let g(x)={[(x^(2)+5x+6)/(x+3)," for "x!=-3],[k," for "x=-3]:} g is continuous for all real numbers. What is the value of k ? Choose 1 answer: (A) -2 (B) -3 (C) -6 (D) -1$

Let $g(x)=\left\{\begin{array}{ll}\frac{x^{2}+5 x+6}{x+3} & \text { for } x \neq-3 \\ k & \text { for } x=-3\end{array}\right.$ $\newline$ $g$ is continuous for all real numbers. $\newline$ What is the value of $k$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $-2$ $\newline$ (B) $-3$ $\newline$ (C) $-6$ $\newline$ (D) $-1$

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Algebra 2

Conjugate root theorems

Full solution

Q. Let

g(x)=\left\{\begin{array}{ll}\frac{x^{2}+5 x+6}{x+3} & \text { for } x \neq-3 \\ k & \text { for } x=-3\end{array}\right.

\newline

g

is continuous for all real numbers.

\newline

What is the value of

k

\newline

Choose

1

answer:

\newline

(A)

-2

\newline

(B)

-3

\newline

(C)

-6

\newline

(D)

-1

Simplifying $g(x)$ : To ensure that $g(x)$ is continuous at $x = -3$ , the limit of $g(x)$ as $x$ approaches $-3$ from the left must equal the value of $g(x)$ at $x = -3$ . Let's first simplify the expression for $g(x)$ when $x$ is not equal to $-3$ . $\newline$ We have: $\newline$ $g(x)$ $1$ $\newline$ This can be factored as: $\newline$ $g(x)$ $2$ $\newline$ For $g(x)$ $3$ , we can cancel out the $g(x)$ $4$ terms: $\newline$ $g(x)$ $5$
Finding the limit of $g(x)$ : Now, we need to find the limit of $g(x)$ as $x$ approaches $-3$ . Since we have simplified $g(x)$ to $x + 2$ for $x \neq -3$ , we can directly substitute $x = -3$ into this simplified expression to find the limit. $\newline$ Limit as $x$ approaches $-3$ of $g(x)$ is: $\newline$ $g(x)$ $1$
Determining the value of $k$ : For $g(x)$ to be continuous at $x = -3$ , the value of $k$ must be equal to the limit of $g(x)$ as $x$ approaches $-3$ . $\newline$ Therefore, $k = -1$ .