In Exercises graphbasicabsvalexerfirst - graphbasicabsvalexerlast, graph the function using Theorem linearabsvaluegraphs. Find the axis intercepts of each graph, if any exist. From the graph, determine the domain and range of each function, the maximum and minimum of each function, if they exist, and list the intervals on which the function is increasing, decreasing or constant.
\(f(x) = |x + 4|\)

Use the Desmos graph below with the settings \(v(x)=|x|,\,h=-4,\,k=0,\,a=1\)

\(x\)-intercept \((-4, 0)\)
\(y\)-intercept \((0, 4)\)
Domain \((-\infty , \infty )\)
Range \([0, \infty )\)
Decreasing on \((-\infty , -4]\)
Increasing on \([-4, \infty )\)
Minimum is \(0\) at \((-4,0)\)
No maximum

\(f(x) = |x| + 4\)

Use the Desmos graph below with the settings \(v(x)=|x|,\,h=0,\,k=4,\,a=1\)

No \(x\)-intercepts
\(y\)-intercept \((0, 4)\)
Domain \((-\infty , \infty )\)
Range \([4, \infty )\)
Decreasing on \((-\infty , 0]\)
Increasing on \([0, \infty )\)
Minimum is \(4\) at \((0,4)\)
No maximum

\(f(x) = |4x|\)

Note \(f(x) = |4x| = 4|x|\).

Use the Desmos graph below with the settings \(v(x)=|x|,\,h=0,\,k=0,\,a=4\)

\(x\)-intercept \((0, 0)\)
\(y\)-intercept \((0, 0)\)
Domain \((-\infty , \infty )\)
Range \([0, \infty )\)
Decreasing on \((-\infty , 0]\)
Increasing on \([0, \infty )\)
Minimum is \(0\) at \((0,0)\)
No maximum

\(g(t) = -3|t|\)

Use the Desmos graph below with the settings \(v(x)=|x|,\,h=0,\,k=0,\,a=-3\)

\(t\)-intercept \((0, 0)\)
\(y\)-intercept \((0, 0)\)
Domain \((-\infty , \infty )\)
Range \((-\infty , 0]\)
Increasing on \((-\infty , 0]\)
Decreasing on \([0, \infty )\)
Maximum is \(0\) at \((0, 0)\)
No minimum

\(g(t) = 3|t + 4| - 4\)

Use the Desmos graph below with the settings \(v(x)=|x|,\,h=-4,\,k=-4,\,a=3\)

\(t\)-intercepts \(\left (-\frac {16}{3}, 0\right )\), \(\left (-\frac {8}{3}, 0\right )\)
\(y\)-intercept \((0, 8)\)
Domain \((-\infty , \infty )\)
Range \([-4, \infty )\)
Decreasing on \((-\infty , -4]\)
Increasing on \([-4, \infty )\)
Minimum is \(-4\) at \((-4,-4)\)
No maximum

\(g(t) = \frac {1}{3}|2t - 1|\)

Note \(g(t) = \frac {1}{3}|2t - 1| = \frac {2}{3}|t - \frac {1}{2}|\)

Use the Desmos graph below with the settings \(v(x)=|x|,\,h=\frac {1}{2},\,k=0,\,a=\frac {2}{3}\)

\(t\)-intercepts \(\left (\frac {1}{2}, 0\right )\)
\(y\)-intercept \(\left (0, \frac {1}{3}\right )\)
Domain \((-\infty , \infty )\)
Range \([0, \infty )\)
Decreasing on \(\left (-\infty , \frac {1}{2}\right ]\)
Increasing on \(\left [\frac {1}{2}, \infty \right )\)
Minimum is \(0\) at \(\left (\frac {1}{2},0\right )\)
No maximum

In Exercises findformulaforabsgraphfirst - findformulaabsgraphlast, find a formula for each function below in the form \(F(x) = a|x-h|+k\).
[Picture]

\(F(x) = 2|x+1|-3\)
[Picture]

\(F(x) = |x-1.25|-2.75\)
[Picture]

\(F(x) = -|x+1|+2\)
[Picture]

\(F(x) = -\frac {1}{2} |x+1|+\frac {3}{2}\)
Use this Desmos link to graph the following pairs of functions on the same set of axes. (The first one is done for you, and you can just modify f(x) to get the others.)
  • \(f(x) = 2-x\) and \(g(x) = | 2-x |\)
  • \(f(x) = x^2-4\) and \(g(x) = | x^2 -4 |\)
  • \(f(x) = x^3\) and \(g(x) = | x^3 |\)
  • \(f(x) = \sqrt {x}-4 \) and \(g(x) = | \sqrt {x} -4| \)

Choose more functions \(f(x)\) and graph \(y = f(x)\) alongside \(y = | f(x)|\) until you can explain how, in general, one would obtain the graph of \(y = | f(x) |\) given the graph of \(y = f(x)\). How does your explanation tie in with with Definition absolutevaluepiecewise?

In each case, the graph of \(g\) can be obtained from the graph of \(f\) by reflecting the portion of the graph of \(f\) which lies below the \(x\)-axis about the \(x\)-axis. This meshes with Definition absolutevaluepiecewise since what we are doing algebraically is making the negative \(y\)-values positive.
Explain why the function below cannot be written in the form \(F(x) = a|x-h|+k\). Write \(F(x)\) as a piecewise-defined linear function.

[Picture]

If \(F(x) = a|x-h| + k\), then for the vertex to be at \((1,-2)\), \(h =1\) and \(k = -2\) so \(F(x) = a |x-1| - 2\). Since \((0,-1)\) is on the graph, \(F(0) = -1\) so \(-1 = a|0-1|-2\) which means \(a = 1\). This means \(F(x) = |x-1|-2\). However, \((2.6,0)\) is also on the graph, so it should work out that \(F(2.6) = 0\). However, we find \(F(2.6) = |2.6-1| - 2 = -0.4 \neq 0\).
\[ F(x) = \begin{cases} -x-1 & \text {if $x \leq 1$, } \\ \frac {5}{4} x - \frac {13}{4} & \text {if $x \geq 1$,} \\ \end{cases}\]
In Exercises graphadvabsvalexerfirst - graphadvabsvalexerlast, graph the function by rewriting each function as a piecewise defined function using Definition absolutevaluepiecewise. Find the axis intercepts of each graph, if any exist. From the graph, determine the domain and range of each function, the maximum and minimum of each function, if they exist, and list the intervals on which the function is increasing, decreasing or constant.
\(f(x) = x + |x| - 3\)

Re-write \(f(x) = x+|x| - 3\) as
\(\displaystyle f(x) = \left \{ \begin{array}{rcl} -3 & \mbox { if } & x < 0\\ 2x -3 & \mbox { if } & x \geq 0 \\ \end{array} \right . \)
\(x\)-intercept \(\left (\frac {3}{2}, 0\right )\)
\(y\)-intercept \((0,-3)\)
Domain \((-\infty , \infty )\)
Range \([-3, \infty )\)
Increasing on \([0,\infty )\)
Constant on \((-\infty , 0]\)
Minimum is \(-3\) at \((x,-3)\) where \(x \leq 0\)
No maximum
\[\graph {x+abs(x)+3}\]
\(f(x) = |x+2| - x\)
\(f(x) = |x+2| - |x|\)
\(g(t) = |t+ 4| + |t- 2|\)
\(g(t) = \frac {|t + 4|}{t + 4}\)
\(g(t) = \frac {|2 - t|}{2 - t}\)
With the help of your classmates, find an absolute value function whose graph is given below.

[Picture]

\(f(x) = | |x|-4|\)
In Exercises solveabsvalequfirsta - solveabsvalequlasta, solve the equation.
\(|x| = 6\)
\(x=-6\) \(x=6\)
\(|3x-1| = 10\)
\(x=-3\) \(x=-\frac {11}{3}\) \(x=0\) \(x=\frac {11}{3}\) \(x=3\)
\(|4-x| = 7\)
\(x=-11\) \(x=-3\) \(x=0\) \(x=3\) \(x=11\)
\(4 - |t| = 3\)
\(t=-7\) \(t=-1\) \(t=0\) \(t=1\) \(t=7\)
\(2|5t+1| - 3 = 0\)
\(t=-\frac {1}{2}\) \(t=-\frac {1}{10}\) \(t=\frac {1}{10}\) \(t=\frac {1}{2}\) No real solutions.
\(|7t-1| + 2 = 0\)
\(t=-\frac {3}{7}\) \(t=-\frac {1}{7}\) \(t=\frac {1}{7}\) \(t=\frac {3}{7}\) No solution
\(\frac {5 - |w|}{2} = 1\)
\(w=-7\) \(w=-3\) \(w=3\) \(w=7\) No solution
\(\frac {2}{3} |5-2w| - \frac {1}{2} = 5\)

\(w = -\frac {13}{8}\) or \(w = \frac {53}{8}\)
\(|w| = w + 3\)

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

\(x=0\) or \(x= 2\)
\(4 - |x| = 2x+1\)

\(x=1\)
\(|x-4| = x-5\)

no solution
Solve the equations in Exercises moreabsvalequfirsta - moreabsvalequlasta using the property that if \(|a| = |b|\) then \(a = \pm b\).
\(|3x - 2| = |2x + 7|\)

\(x = -1\) or \(x = 9\)
\(|3x+1| = |4x|\)

\(x = -\frac {1}{7}\) or \(x = 1\)
\(|1-2x| = |x+1|\)

\(x = 0\) or \(x = 2\)
\(|4-t| - |t+2| = 0\)

\(t=1\)
\(|2-5t| = 5 |t+1|\)

\(t = -\frac {3}{10}\)
\(3|t-1| = 2|t+1|\)

\(t = \frac {1}{5}\) or \(t = 5\)
In Exercises solveinequabsfirst - solveinequabslast, solve the inequality. Write your answer using interval notation.
\(|3x - 5| \leq 4\)

\(\left [\frac {1}{3}, 3\right ]\)
\(|7x + 2| > 10\)

\(\left (-\infty , -\frac {12}{7} \right ) \cup \left (\frac {8}{7}, \infty \right )\)
\(|2t+1| - 5 < 0\)

\((-3,2)\)
\(|2-t| - 4 \geq -3\)

\((-\infty ,1] \cup [3,\infty )\)
\(|3w+5| + 2 < 1\)

No solution
\(2|7-w| +4 > 1\)

\((-\infty , \infty )\)
\(2 \leq |4-x| < 7\)

\((-3,2] \cup [6,11)\)
\(1 < |2x - 9| \leq 3\)

\([3, 4) \cup (5, 6]\)
\(|t + 3| \geq |6t + 9|\)

\(\left [-\frac {12}{7}, -\frac {6}{5}\right ]\)
\(|t-3| - |2t+1| < 0\)

\((-\infty , -4) \cup \left ( \frac {2}{3}, \infty \right )\)
\(|1-2x| \geq x + 5\)

\(\left (-\infty , -\frac {4}{3} \right ] \cup [6, \infty )\)
\(x + 5 < |x+5|\)

\((-\infty , -5)\)
\(x \geq |x+1|\)

No Solution.
\(|2x + 1| \leq 6-x\)

\(\left [ -7, \frac {5}{3}\right ]\)
\(t + |2t-3| < 2\)

\(\left ( 1, \frac {5}{3} \right )\)
\(|3-t| \geq t-5\)

\((-\infty , \infty )\)
Show that if \(\delta \) is a real number with \(\delta > 0\), the solution to \(|x-a| < \delta \) is the interval: \((a - \delta , a + \delta )\). That is, an interval centered at \(a\) with ‘radius’ \(\delta \).
The Triangle Inequality for real numbers states that for all real numbers \(x\) and \(a\), \(|x+a| \leq |x| + |a|\) and, moreover, \(|x+a| = |x|+|a|\) if and only if \(x\) and \(a\) are both positive, both negative, or one or the other is \(0\). Graph each pair of functions below on the same pair of axes and use the graphs to verify the triangle inequality in each instance.
  • \(f(x) = |x+2|\) and \(g(x) = |x|+2\).
  • \(f(x) = |x+4|\) and \(g(x) = |x|+4\).