Consider the function \(g(x)=-\dfrac{12}{x}\)
- Is \(g(x)\) an odd or even function? Give reason(s) for your answer. (2 marks)
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- Sketch the graph \(y=g(x)\) in the domain \(-6 \leq x \leq 6\). Label the endpoints and two other points on the curve. (2 marks)
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a. \(\text{If}\ g(x)\ \text{is odd}\ \ \Rightarrow \ \ g(-x)=-g(x)\)
\(-g(x)= \dfrac{12}{x}\)
\(g(-x) = -\dfrac{12}{(-x)} = \dfrac{12}{x} = -g(x)\)
\(\therefore g(x)\ \text{is odd (and cannot be even).}\)
b. \(\text{Table of values:}\)
\begin{array}{|c|c|c|c|c|c|}
\hline \rule{0pt}{2.5ex}\ \ x \ \ \rule[-1ex]{0pt}{0pt}& -4 & -2 & \ \ 0 \ \ & 2 & 4 \\
\hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& 3 & 6 & \infty & -6 & -3 \\
\hline
\end{array}
\(\text{Endpoints:}\ (-6,2), (6,-2) \)
a. \(\text{If}\ g(x)\ \text{is odd}\ \ \Rightarrow \ \ g(-x)=-g(x)\)
\(-g(x)= \dfrac{12}{x}\)
\(g(-x) = -\dfrac{12}{(-x)} = \dfrac{12}{x} = -g(x)\)
\(\therefore g(x)\ \text{is odd.}\)
b. \(\text{Table of values:}\)
\begin{array}{|c|c|c|c|c|c|}
\hline \rule{0pt}{2.5ex}\ \ x \ \ \rule[-1ex]{0pt}{0pt}& -4 & -2 & \ \ 0 \ \ & 2 & 4 \\
\hline \rule{0pt}{2.5ex}y \rule[-1ex]{0pt}{0pt}& 3 & 6 & \infty & -6 & -3 \\
\hline
\end{array}
\(\text{Endpoints:}\ (-6,2), (6,-2) \)