Exercise 1.1.49 (49-52)

Question

Evaluate $f(-3), f(0)$, and $f(2)$ for the piecewise defined function. Then sketch the graph of the function. ewline 49. $f(x)= \begin{cases}x^2+2 & ext { if } x<0 \ x & ext { if } x \geqslant 0\end{cases}$ ewline 50. $f(x)= \begin{cases}5 & ext { if } x<2 \ \frac{1}{2} x-3 & ext { if } x \geqslant 2\end{cases}$ ewline 51. $f(x)= \begin{cases}x+1 & ext { if } x \leqslant-1 \ x^2 & ext { if } x>-1\end{cases}$ ewline 52. $f(x)= \begin{cases}-1 & ext { if } x \leqslant 1 \ 7-2 x & ext { if } x>1\end{cases}$

Khagan Gulusoy · Accepted Answer

\begin{enumerate}

\item[49.] \begin{itemize}
    \item 	extit{(evaluation)}$$
\begin{aligned}
& f(-3)=(-3)^2+2=9+2=11 \
& f(0)=0 \
& f(2)=2
\end{aligned}
$$
\item 	extit{(graph)} To sketch the graph, we need first define this function numerically, i.e., by a table of values.
\begin{table}[H]\centering
\begin{tabular}{|l|l|l|l|l|l|l|}
\cline{1-4} \cline{6-7}
$x$ & $-2$ & $-1$ & $-3$ &  & $0$ & $1$ \ \cline{1-4} \cline{6-7} 
$y$ & $6$  & $3$  & $11$ &  & $0$ & $1$ \ \cline{1-4} \cline{6-7} 
\end{tabular}
\end{table}
After we connect the points, we get:
\end{itemize}
\begin{figure}[H] \centering
\begin{tikzpicture}
    \begin{axis}[axis lines=middle,xmin=-4.2,xmax=2.2,ymin=-1,ymax=12.2,xlabel={$x$}, ylabel={$f(x)$}, x label style = {at={(ticklabel cs:0.9)},anchor=north, above=6.7mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-4:0, samples=150, mark=none, color=red, ultra thick] {x^2+2}; 
         \addplot[domain=0:2, samples=2, mark=none, color=red, ultra thick] {x};          
         \addplot[mark=*, only marks] coordinates {(-3,11) (-2,6) (-1, 3)}; 
         \addplot[mark=*, only marks] coordinates {(0,0) (1, 1)}; 
         \addplot[mark=*, fill=white, only marks] coordinates {(0, 2)}; 
         
    \end{axis}
\end{tikzpicture}
\end{figure}

\item[50.]\begin{itemize}
    \item 	extit{(evaluation)} $$
\begin{aligned} 
& f(-3)=5 \
& f(0)=5 ; \
& f(2)=\frac{1}{2} \cdot 2-3=-2
\end{aligned}
$$
\item 	extit{(graph)} To sketch the graph, we need first define this function numerically, i.e., by a table of values.
\begin{table}[H]\centering
\begin{tabular}{|l|l|l|l|l|l|}
\cline{1-3} \cline{5-6}
$x$ & $0.5$ & $1$ &  & $3$    & $4$  \ \cline{1-3} \cline{5-6} 
$y$ & $5$   & $5$ &  & $-1.5$ & $-1$ \ \cline{1-3} \cline{5-6} 
\end{tabular}
\end{table}
Now we can easily sketch the graph:
\begin{figure}[H] \centering
\begin{tikzpicture}
    \begin{axis}[axis lines=middle,xmin=-1,xmax=4.2,ymin=-2.2,ymax=5.2,xlabel={$x$}, ylabel={$f(x)$}, x label style = {at={(ticklabel cs:0.9)},anchor=north, above=6.7mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-1:2, samples=2, mark=none, color=violet, ultra thick] {5}; 
         \addplot[domain=2:4.2, samples=2, mark=none, color=violet, ultra thick] {1/2*x-3};          
         \addplot[mark=*, only marks] coordinates {(3,-1.5) (4,-1)}; 
         \addplot[mark=*, only marks] coordinates {(-0.5,5) (1, 5) (2,-2)}; 
         \addplot[mark=*, fill=white, only marks] coordinates {(2, 5)}; 
         
    \end{axis}
\end{tikzpicture}
\end{figure}
\end{itemize}

\item[51.] \begin{itemize}
    \item (	extit{evaluation}) $$
\begin{aligned}
& f(-3)=-3+1=-2 ; \
& f(0)=0^2=0 ; \
& f(2)=2^2=4
\end{aligned}
$$

\item (	extit{graph}) As always, we first create the table:
\begin{table}[H]\centering
\begin{tabular}{|l|l|l|l|l|l|l|l|}
\cline{1-1} \cline{3-4} \cline{6-8}
$x$ &  & $-1.5$ & $-1$ &  & $-0.5$ & $0$ & $1$ \ \cline{1-1} \cline{3-4} \cline{6-8} 
$y$ &  & $-0.5$ & $0$  &  & $0.25$ & $0$ & $1$ \ \cline{1-1} \cline{3-4} \cline{6-8} 
\end{tabular}
\end{table}
Now we are able to sketch the graph:
\begin{figure}[H] \centering
\begin{tikzpicture}
    \begin{axis}[axis lines=middle,xmin=-2,xmax=1.2,ymin=-1.2,ymax=2.2,xlabel={$x$}, ylabel={$f(x)$}, x label style = {at={(ticklabel cs:0.9)},anchor=north, above=6.7mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-2:-1, samples=2, mark=none, color=blue, ultra thick] {x+1}; 
         \addplot[domain=-1:1.2, samples=150, mark=none, color=blue, ultra thick] {x^2};          
         \addplot[mark=*, only marks] coordinates {(-0.5,0.25) (0,0) (1,1)}; 
         \addplot[mark=*, only marks] coordinates {(-1.5,-0.5) (-1, 0)}; 
         \addplot[mark=*, fill=white, only marks] coordinates {(-1, 1)}; 
         
    \end{axis}
\end{tikzpicture}
\end{figure}

\end{itemize}

\item[52.]
\begin{itemize}
    \item 	extit{(evaluation)} $$
\begin{aligned}
& f(-3)=-1 ; \
& f(0)=-1 ; \
& f(2)=7-2 \cdot 2=7-4=3
\end{aligned}
$$

\item 	extit{(graph)} To sketch the graph, we need first define this function numerically, i.e., by a table of values.
\begin{table}[H]\centering
\begin{tabular}{|l|l|c|c|l|l|l|}
\cline{1-1} \cline{3-4} \cline{6-7}
$x$ &  & $-1$ & $1$  &  & $2$ & $3$ \ \cline{1-1} \cline{3-4} \cline{6-7} 
$y$ &  & $-1$ & $-1$ &  & $3$ & $1$ \ \cline{1-1} \cline{3-4} \cline{6-7} 
\end{tabular}
\end{table}
Now we can sketch the graph:
\begin{figure}[H] \centering
\begin{tikzpicture}
    \begin{axis}[axis lines=middle,xmin=-1.2,xmax=3.2,ymin=-1.2,ymax=5.2, ytick={-1, 1, 3, 5}, xlabel={$x$}, ylabel={$f(x)$}, x label style = {at={(ticklabel cs:0.9)},anchor=north, above=6.7mm}, y label style={anchor=north, above=3mm}]
         \addplot[domain=-1.2:1, samples=2, mark=none, color=black!60!green, ultra thick] {-1}; 
         \addplot[domain=1:3.2, samples=2, mark=none, color=black!60!green, ultra thick] {7-2*x};          
         \addplot[mark=*, only marks] coordinates {(2,3) (3,1)}; 
         \addplot[mark=*, only marks] coordinates {(-1,-1) (1, -1)}; 
         \addplot[mark=*, fill=white, only marks] coordinates {(1, 5)}; 
         
    \end{axis}
\end{tikzpicture}
\end{figure}

\end{itemize}

\end{enumerate}

Exercise 1.1.49 (49-52)

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Exercise 1.1.49 (49-52)

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