Let
\begin{equation*}
f(x) = y = \sqrt{x^2-4}
\end{equation*}
For Domain:
\begin{align*}
x^2-4 \amp \geq 0 \\
(x+2)(x-2) \amp \geq 0 \\
|x| \amp \geq \pm 2 \\
i.e., \, -2 \leq x \amp \geq 2
\end{align*}
Hence, the domian: \((-\infty, -2] \cup [2, \infty)\) (Use number line for details understanding)
For Range:
\begin{align*}
y \amp = \sqrt{x^2-4} \\
or, \, y^2 \amp =x^2-4 \\
or, \, x \amp = \sqrt{y^2+4}
\end{align*}
For \(y\) to be \(+ve\) the Range: \([0, \infty) \) As \(x\to -\infty \) \(y \to \infty\) and as \(x\to \pm 2 \) \(y \to 0\)
For Inverse: Interchange \(y\) and \(x.\)
\begin{equation*}
x = \sqrt{y^2-4}
\end{equation*}
\begin{equation*}
or, \, x^2 = y^2-4
\end{equation*}
\begin{equation*}
or, \, y =\sqrt{x^2+4} =f^{-1}(x)
\end{equation*}
To draw graph of a function: determine its shape:
-
-
-
standard plot of
\(y=\sqrt{x}\)
Find Point (x,y): set
\begin{equation*}
x^2-4=0
\end{equation*}
\begin{equation*}
or, \, x=\pm 2
\end{equation*}
\begin{equation*}
or, \, y = \sqrt{4-4} = 0
\end{equation*}
Hence, point (x,y) =
\((-2, 0)\) and
\((2, 0)\)
Intercept:
-
y-intercept: set \(x=0 \)
\begin{equation*}
y = \sqrt{-4} = undefined
\end{equation*}
-
x-intercept: set \(y=0 \)
\begin{equation*}
x=\pm 2
\end{equation*}
The final graph looks like:
