Solution.

Recall the domain of the linear function is \(\mathbb {R}\) the set of real numbers. Therefore to get two points on the line we use any two real numbers for \(x\),

When \(x=0\implies y=3\). \((0,3)\) is one point on the line.
When \(y=0\implies x=\frac {3}{2}\). \(\left (\frac {3}{2},0\right )\) is another point.

Solution.

This function divides the domain into parts for \(-\infty < x<2,\) \(y=-\frac {1}{2}x+1\) which is a straight line.
Also for \(2\leq x< \infty ,\) \(y=\frac {1}{2}x-1\) which is also a straight line.
Therefore we need to sketch two straight lines for this graph.

\(-\infty <x<2,\) \(y=-\frac {1}{2}x+1\)

when \(x=0,\) \(y=1\)
when \(y=0,\) \(x=2\)

For \(2\leq x < \infty ,\) \(y=\frac {1}{2}x-1\)

when \(x=2,\) \(y=0\)

when \(x=4,\) \(y=1\)

Thus

This is the graph of \( y= \begin {cases} -\frac {1}{2}x+1, & x<2\\\\ \frac {1}{2}x-1, & x\geq 2\\ \end {cases} \)

And this is exactly the graph of \(y=\bigg |\frac {1}{2}x-1\bigg |\), because that absolute value unpacks into the very piecewise function just sketched:

\begin {align*} y=\bigg |\frac {1}{2}x-1\bigg | & = \begin {cases} \frac {1}{2}x-1 & \text {if}\quad \frac {1}{2}x-1\geq 0\\\\ -\left (\frac {1}{2}x-1\right ) & \text {if}\quad \frac {1}{2}x-1<0\\ \end {cases}\\\\ &= \begin {cases} \frac {1}{2}x-1 & \text {if}\quad x\geq 2\\\\ -\frac {1}{2}x+1 & \text {if}\quad x<2\\ \end {cases}\\ \end {align*}

Recall that \(|x|= \begin {cases} x & \text {if}\quad x\geq 0\\ -x & \text {if}\quad x<0\\ \end {cases} \)

Thus to sketch the graph of \(y=|x|\), we use the definition to write the function as \[y= \begin {cases} x & \text {if}\quad x\geq 0\\ -x & \text {if}\quad x<0\\ \end {cases} \]

This gives two straight lines whose graphs we can easily sketch as follows

For \(x<0,\) \(y=-x\)					For \(x\geq 0,\) \(y=x\)

combining the two graphs we obtain the following graph

Solution.

Using the definition \begin {align*} f(x)=|3x-3| & = \begin {cases} 3x-3 & \text {if}\quad 3x-3\geq 0\\ -(3x-3) & \text {if}\quad 3x-3<0\\ \end {cases}\\\\ \implies \quad f(x) & = \begin {cases} 3x-3 & \text {if}\quad x\geq 1\\ -3x+3 & \text {if}\quad x<1 \end {cases} \end {align*}

Solution.

\(f(x)= \begin {cases} x^2+x-6 & \text {if}\quad x^2+x-6\geq 0\\\\ -(x^2+x-6) & \text {if}\quad x^2+x-6<0\\ \end {cases} \)

\(x^2+x-6\geq 0\implies (x+3)(x-2)\geq 0\) so the critical values are \(x=-3\) and \(x=2\)

• \(x^2+x-6\geq 0\) in the interval \((-\infty ,-3]\cup [2,\infty )\)
• \(x^2+x-6<0\) in the interval \((-3,2)\)
  

\(f(x)= \begin {cases} x^2+x-6 & \text {for}\quad (-\infty ,-3]\cup [2,\infty )\\\\ -x^2-x+6 & \text {for}\quad (-3,2)\\ \end {cases} \)

Solution.

The graph of \(f(x)=-|2x-1|\) is the graph of \(g(x)=|2x-1|\) but reflected in the \(x-\)axis.

Solution.

(a)

\(f(x)=|x+3|-2\).
First consider the graph of \(h(x)=|x+3|\).

(b)

\(f(x)=1-|2x+1|\)
First consider the graph of \(g(x)=|2x+1|\). Then sketch the graph of
\(h(x)=-|2x+1|\) which is a reflection of the graph of \(g(x)=|2x+1|\) in the \(x-\)axis.

(c)

\(f(x)=|2x+3|-|3x-3|\)
Here we use the definition to find the various straight lines to sketch. The critical values are where each bracket changes sign: \(2x+3=0\) at \(x=-\frac {3}{2}\), and \(3x-3=0\) at \(x=1\).

So we redefine the function in the intervals \[\bigg (-\infty ,-\frac {3}{2}\bigg ),\quad \bigg [-\frac {3}{2},1\bigg ),\quad [1,\infty )\]

	\((-2)\)	\(-\frac {3}{2}\)	0	1	\((2)\)
\(2x+3\)	\(-\)	\(|\)	\(+\)	\(|\)	\(+\)
\(3x-3\)	\(-\)	\(|\)	\(-\)	\(|\)	\(+\)

Below \(-\frac {3}{2}\) both brackets are negative; between \(-\frac {3}{2}\) and \(1\) the first is positive and the second still negative; above \(1\) both are positive. Therefore \[f(x)= \begin {cases} -(2x+3)+(3x-3) & \text {for}\quad x<-\frac {3}{2}\\[4pt] (2x+3)+(3x-3) &\text {for}\quad -\frac {3}{2}\leq x<1\\[4pt] (2x+3)-(3x-3) &\text {for}\quad x\geq 1 \end {cases}\]

Now

i.

\(-(2x+3)+(3x-3)=-2x-3+3x-3=x-6\)

ii.

\((2x+3)-(-)(3x-3)=2x+3+3x-3=5x\)

iii.

\((2x+3)-(3x-3)=2x+3-3x+3=-x+6\)

The function becomes \(f(x)= \begin {cases} x-6 & \text {if}\quad x<-\frac {3}{2}\\\\ 5x & \text {if} \quad -\frac {3}{2}\leq x<1\\\\ -x+6 & \text {if}\quad x\geq 1\\ \end {cases} \)