Exercise 1.1.47

Question

Find the domain and range and sketch the graph of the function $h(x)=\sqrt{4-x^2}$.

Khagan Gulusoy · Accepted Answer

\begin{itemize}
\item (	extit{domain}) Since the term under any even-powered root must be bigger than or equal to zero, we have: 
    \begin{align*}
        \forall x \in \mathbb{R}:\quad 
        &4-x^2\geq 0\
        &\iff (2-x)(2+x) \geq 0\
        &\iff (x-2)(x+2)\leq 0\
        &\iff x\geq -2 	ext{ and } x \leq 2.
    \end{align*}
    In other words, the domain of $h$ is $\{x \in \mathbb{R} \mid x \geq -2 	ext{ and } x \leq 2\} = [-2, \ 2]$.
\item (	extit{range}) 
    The minimum output of this function is $0$. The function reaches its maximum when the term under the root is maximal. That is why the maximum value of $h$ is equal to $\sqrt{4-0^2}=\sqrt{4}=2.$ In other words, the range of $h$ is $\{h(x) \in \mathbb{R} \mid x\in [-2,2]\}=[0, 2].$

\item (	extit{graph}) To sketch the graph, we first need to define this function numerically, i.e., by a table of values.
    \begin{table}[H]\centering
\begin{tabular}{|l|l|l|l|l|l|}
\hline
$x$ & $-2$ & $0$ & $2$ & $-1$       & $1$        \ \hline
$y$ & $0$  & $2$ & $0$ & $\sqrt{3}$ & $\sqrt{3}$ \ \hline
\end{tabular}
\end{table}
    We smoothly connect the points from the table and sketch the graph of $h(x)=\sqrt{4-x^2}$ as follows (we use the fact that $\sqrt{3}\approx 1.7)$:
    \begin{figure}[H] \centering
\begin{tikzpicture}
    \begin{axis}[axis lines=middle,width=6cm, height=4cm,xmin=-2.2,xmax=2.2,ymin=0,ymax=2.2,xlabel={$x$}, ylabel={$h(x)$}, x label style = {at={(ticklabel cs:0.5)},anchor=north}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-2:2, samples=150, mark=none, color=violet, ultra thick] {sqrt(4-x^2)}; 
         \addplot[mark=*, only marks] coordinates {(-2, 0) (-1, 1.73205) (0,2) (1, 1.73205) (2,0)}; 
    \end{axis}
\end{tikzpicture}
\end{figure}
\end{itemize}

Exercise 1.1.47

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Exercise 1.1.47

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