Find the graph of the following functions.
f\left(x\right)=\left| 3-2x \right|
We know that:
\left| 3-2x \right|=\begin{cases} 3-2x \text{ if }3-2x\geq0 \cr \cr -\left(3-2x\right) \text{ if }3-2x\leq0 \end{cases}
We have:
3-2x \ge0 \Leftrightarrow 2x \ge 3 \Leftrightarrow x \ge \dfrac{3}{2}
And:
3-2x \leq 0 \Leftrightarrow 2x \leq 3 \Leftrightarrow x \leq \dfrac{3}{2}
Therefore:
\left| 3-2x \right|=\begin{cases} 3-2x \text{ if }x\geq \dfrac{3}{2} \cr \cr 2x-3 \text{ if }x\leq \dfrac{3}{2} \end{cases}
We draw the two lines to obtain the graph of f:
The graph of f is as follows:
f\left(x\right)=-\left| x+1 \right|
We know that:
-\left| x+1 \right|=\begin{cases} -\left(x+1\right) \text{ if }x+1\geq0 \cr \cr -\left(-\left(x+1\right)\right) \text{ if }x+1\leq0 \end{cases}
We have:
x+1 \ge0 \Leftrightarrow x \ge -1
And:
x+1 \lt 0 \Leftrightarrow x \lt -1
Therefore:
-\left| x+1 \right|=\begin{cases} -x-1 \text{ if }x\geq -1 \cr \cr x+1 \text{ if }x\leq -1 \end{cases}
We draw the two lines to obtain the graph of f:
The graph of f is as follows:
f\left(x\right)=1-|2x-3|
We know that:
1-\left| 2x-3 \right|=\begin{cases}1-2x+3 \text{ if }2x-3\geq0 \cr \cr 1-\left(-\left(2x-3\right)\right) \text{ if }2x-3\leq0 \end{cases}
We have:
2x-3 \ge0 \Leftrightarrow 2x \ge 3 \Leftrightarrow x \ge \dfrac{3}{2}
And:
2x -3\lt 0 \Leftrightarrow 2x \lt 3 \Leftrightarrow x \lt \dfrac{3}{2}
Therefore:
1-\left| 2x-3 \right|=\begin{cases} 4-2x \text{ if }x\geq \dfrac{3}{2} \cr \cr 2x-2 \text{ if }x\leq \dfrac{3}{2} \end{cases}
We draw the two lines to obtain the graph of f:
The graph of f is as follows:
f\left(x\right)=x+|x|+1
We know that:
x+|x|+1=\begin{cases} x+x+1 \text{ if }x\geq0 \cr \cr x-x+1 \text{ if }x\leq0 \end{cases}
Therefore:
x+|x|+1=\begin{cases}2x+1 \text{ if }x\geq0 \cr \cr1 \text{ if }x\leq0 \end{cases}
We draw the two lines to obtain the graph of f:
The graph of f is as follows:
f\left(x\right)=x + |2x-1|-\dfrac{1}{2}
We know that:
x + |2x-1|-\dfrac{1}{2}=\begin{cases} x+2x-1-\dfrac{1}{2} \text{ if }2x-1\geq0 \cr \cr x-2x+1-\dfrac{1}{2} \text{ if }3-2x\leq0 \end{cases}
We have:
2x-1 \ge0 \Leftrightarrow 2x \ge 1 \Leftrightarrow x \ge \dfrac{1}{2}
And:
2x-1 \lt 0 \Leftrightarrow 2x \lt 1 \Leftrightarrow x \lt \dfrac{1}{2}
Therefore:
x + |2x-1|-\dfrac{1}{2}=\begin{cases} 3x-\dfrac{3}{2} \text{ if }x\geq \dfrac{1}{2}\cr \cr -x+\dfrac{1}{2} \text{ if }x \lt \dfrac{1}{2} \end{cases}
We draw the two lines to obtain the graph of f:
The graph of f is as follows:
f\left(x\right)=|x|+|x-1|
We know that:
|x|=\begin{cases} x\text{ if }x\geq0 \cr \cr -x \text{ if }x \lt 0 \end{cases}
And:
|x-1|=\begin{cases} x-1\text{ if }x\geq 1\cr \cr -x+1 \text{ if }x \lt 1 \end{cases}
Therefore:
|x|+|x-1|=\begin{cases} x + \left(x-1\right) \text{ if }x\geq 1 \cr \cr x -\left(x-1\right) \text{ if } 0 \le x \lt 1 \cr \cr -x -\left(x-1\right) \text{ if }x \lt 0 \end{cases}
|x|+|x-1|=\begin{cases}2x-1 \text{ if }x\geq 1 \cr \cr 1 \text{ if } 0 \le x \lt 1 \cr \cr -2x+1\text{ if }x \lt 0 \end{cases}
We draw the three lines to obtain the graph of f:
The graph of f is as follows:
f\left(x\right)=|x|-|x-1|
We know that:
|x|=\begin{cases} x\text{ if }x\geq0 \cr \cr -x \text{ if }x \lt 0 \end{cases}
And:
|x-1|=\begin{cases} x-1\text{ if }x\geq 1\cr \cr -x+1 \text{ if }x \lt 1 \end{cases}
Therefore:
|x|+|x-1|=\begin{cases} x - \left(x-1\right) \text{ if }x\geq 1 \cr \cr x -\left[-\left(x-1\right)\right] \text{ if } 0 \le x \lt 1 \cr \cr -x -\left[-\left(x-1\right)\right] \text{ if }x \lt 0 \end{cases}
|x|+|x-1|=\begin{cases}1 \text{ if }x\geq 1 \cr \cr 2x-1 \text{ if } 0 \le x \lt 1 \cr \cr -1\text{ if }x \lt 0 \end{cases}
We draw the three lines to obtain the graph of f:
The graph of f is as follows:
