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g(x)={[(5)/(x+3)," if "x >= 0],[(5)/(x-3)," if "x < 0]:}
What is the domain of function
g ?
Choose 1 answer:
(A)
{x inR}
(B)
{x inR∣x!=+-3}
(C)
{x inR∣-3 < x < 3}
(D)
{x inR∣x < -3 or
x > 3}

$g(x)=\left\{\begin{array}{ll} \frac{5}{x+3} & \text { if } x \geq 0 \\ \frac{5}{x-3} & \text { if } x<0 \end{array}\right.$ $\newline$ What is the domain of function $g$ ? $\newline$ Choose $1$ answer: $\newline$ (A) $\{x \in \mathbb{R}\}$ $\newline$ (B) $\{x \in \mathbb{R} \mid x \neq \pm 3\}$ $\newline$ (C) $\{x \in \mathbb{R} \mid-3<x<3\}$ $\newline$ (D) $\{x \in \mathbb{R} \mid x<-3$ or $x>3\}$

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Intermediate Value Theorem

Full solution

Q.

g(x)=\left\{\begin{array}{ll} \frac{5}{x+3} & \text { if } x \geq 0 \\ \frac{5}{x-3} & \text { if } x<0 \end{array}\right.

\newline

What is the domain of function

g

?

\newline

Choose

1

answer:

\newline

(A)

\{x \in \mathbb{R}\}

\newline

(B)

\{x \in \mathbb{R} \mid x \neq \pm 3\}

\newline

(C)

\{x \in \mathbb{R} \mid-3<x<3\}

\newline

(D)

\{x \in \mathbb{R} \mid x<-3

or

x>3\}

Identify Denominator Restrictions: For $x \geq 0$ , $g(x) = \frac{5}{(x+3)}$ . We can't have a denominator of zero, so $x+3$ cannot be zero. This means $x$ cannot be $-3$ .
Define Function for $x \geq 0$ : For $x < 0$ , $g(x) = \frac{5}{(x-3)}$ . Similarly, $x-3$ cannot be zero, so $x$ cannot be $3$ .
Define Function for $x < 0$ : Combining both conditions, the domain of $g(x)$ is all real numbers except $x$ cannot be $-3$ or $3$ .
Combine Conditions for Domain: The correct answer is (B) ${x \in \mathbb{R} | x \neq \pm 3}$ .

More problems from Intermediate Value Theorem

Question

Which of the following functions are continuous for all real numbers?

\newline

g(x)=\ln (x)

\newline

f(x)=\frac{1}{x}

\newline

1

\newline

g

\newline

f

\newline

g

f

\newline

g

f

Posted 3 months ago

Question

\newline

f(x)=\left\{\begin{array}{ll} \ln (-x)+3 & \text { for } x<-3 \\ \ln (-x+3) & \text { for }-3 \leq x<3 \end{array}\right.

\newline

f

x=-3

\newline

1

\newline

\newline

\mathrm{No}

Posted 2 months ago

Question

\newline

f(x)=\left\{\begin{array}{ll}\ln (x) & \text { for } 0<x \leq 2 \\ x^{2} \ln (2) & \text { for } x>2\end{array}\right.

\newline

f

x=2

\newline

1

\newline

\newline

\mathrm{No}

Posted 3 months ago

Question

\newline

g(x)=\left\{\begin{array}{ll} \frac{1}{\cos (x)} & \text { for }-\frac{\pi}{2}<x<0 \\ \cos (x+\pi) & \text { for } x \geq 0 \end{array}\right.

\newline

g

x=0

\newline

1

\newline

\newline

Posted 3 months ago

Question

\newline

h(x)=\left\{\begin{array}{ll} 2^{x-1} & \text { for } x<1 \\ 2^{1-x} & \text { for } x \geq 1 \end{array}\right.

\newline

h

x=1

\newline

1

\newline

\newline

Posted 3 months ago

Question

\newline

g(x)=\left\{\begin{array}{ll} (x-2)^{2} & \text { for } x \leq 2 \\ 2-x^{2} & \text { for } x>2 \end{array}\right.

\newline

g

x=2

\newline

1

\newline

\newline

\mathrm{No}

Posted 3 months ago

Question

\newline

g(x)=\left\{\begin{array}{ll} e^{x} & \text { for } x \leq-1 \\ -e^{-x} & \text { for } x>-1 \end{array}\right.

\newline

g

x=-1

\newline

1

\newline

\newline

Posted 3 months ago

Question

\newline

h(x)=\left\{\begin{array}{ll} e^{2 x} & \text { for } x<0 \\ e^{5 x} & \text { for } x \geq 0 \end{array}\right.

\newline

h

x=0

\newline

1

\newline

\newline

Posted 3 months ago

Question

\newline

h(x)=\left\{\begin{array}{cl}\frac{6}{x+5} & \text { for }-5<x<-3 \\ x^{2}-6 & \text { for } x \geq-3\end{array}\right.

\newline

h

x=-3

\newline

1

\newline

\newline

Posted 3 months ago

Question

\newline

g(x)=\left\{\begin{array}{ll}\sin (x) & \text { for } x<0 \\ x^{2} & \text { for } x \geq 0\end{array}\right.

\newline

g

x=0

\newline

1

\newline

\newline

Posted 3 months ago