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$Let h(x)={[(3x)/(xe^(x))," for "x!=0],[k," for "x=0]:} h is continuous for all real numbers. What is the value of k ? Choose 1 answer: (A) 1 (B) 3 (C) 0 (D) e$

Let $h(x)=\left\{\begin{array}{ll}\frac{3 x}{x e^{x}} & \text { for } x \neq 0 \\ k & \text { for } x=0\end{array}\right.$ $\newline$ $h$ is continuous for all real numbers. $\newline$ What is the value of $k$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $1$ $\newline$ (B) $3$ $\newline$ (C) $0$ $\newline$ (D) $e$

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Algebra 2

Conjugate root theorems

Full solution

Q. Let

h(x)=\left\{\begin{array}{ll}\frac{3 x}{x e^{x}} & \text { for } x \neq 0 \\ k & \text { for } x=0\end{array}\right.

\newline

h

is continuous for all real numbers.

\newline

What is the value of

k

\newline

Choose

1

answer:

\newline

(A)

1

\newline

(B)

3

\newline

(C)

0

\newline

(D)

e

Determine the value of $k$ : To determine the value of $k$ that makes $h(x)$ continuous at $x = 0$ , we need to find the limit of $h(x)$ as $x$ approaches $0$ from the left and right and set it equal to $h(0)$ . The function is given by: $\newline$ $h(x) = \begin{cases} \frac{3x}{xe^{x}} & \text{for } x \neq 0,\ k & \text{for } x = 0 \end{cases}$ $\newline$ First, we simplify the expression for $x \neq 0$ : $\newline$ $h(x) = \frac{3}{e^{x}}$ $\newline$ Now, we find the limit as $x$ approaches $0$ : $\newline$ $\lim_{x \to 0} h(x) = \lim_{x \to 0} \left(\frac{3}{e^{x}}\right)$ $\newline$ Since $k$ $1$ , the limit is: $\newline$ $\lim_{x \to 0} h(x) = \frac{3}{1} = 3$
Simplify the expression: Since $h(x)$ must be continuous at $x = 0$ , the value of $h(0)$ must be equal to the limit we just found. Therefore, we set $k$ equal to the limit: $\newline$ $k = \lim_{x \to 0} h(x) = 3$
Find the limit as $x$ approaches $0$ : We have determined that for $h(x)$ to be continuous at $x = 0$ , $k$ must be $3$ . This corresponds to answer choice (B).