Absolute value functions

The function with absolute value is defined as $$ f(x) = |x| = \begin{cases}x& \quad \mathrm{if}\quad x \geq 0\\ -x& \quad \mathrm{if}\quad x < 0.\end{cases}$$ Function domain is the set of real numbers. The graph shown at following figure consists of two half-lines. For positive values of \(x\) the graph is half-line which is a part of the function \(y=x\) while for negative values of \(x\) the graph is half-line which is part of the function \(y=-x.\)

The function is located above the x axis except for \(x=0\), which means that \(|x|\leq 0\) for all real numbers.The function is even \(f(-x)=f (x).\) The graph is symmetrical with respect to the y-axis. From \(-infty\) to 0 function decreases while from 0 to \(+infty\) the function increases. In origin the function has the minimum value (the value 0).

• Example 1 Draw the function \(f(x) = |3x-2| \)
  Solution First, let's find the zero of a function \(g(x) = 3x-2\). The term "zero of the function" means to find the value of the argument \(x\) for which the function gives zero value. \begin{eqnarray} g(x) &=& 0 \\ 3x-2 &=& 0 \\ 3x &=& 2 /:3 \\ x&=& \frac{2}{3}\end{eqnarray} The zero of the function \(g(x) = 3x -2 \) is \(\left(\frac{2}{3}, 0\right)\). Both curves \(f(x)\),and \(g(x)\) are shown in the following figure.
  
  
  
  
  
  Those two functions shown in previous figure are shown to see the difference between the linear function \(g(x)\) and the function \(f(x)\) with absolute value. As seen from figure the function \(g(x)\) can be divided on two segements that are divided using zero of the function calculated previously. The first segment is for \(x\) in range from \(-\infty\) to \(\frac{2}{3}\). In this segment the values of the function are negative. In the second segment \(x\in \Big[\frac{2}{3}, +\infty \Big\rangle\) the function values are positive. To transform \(g(x)\) into \(f(x)\) the line segment that is under \(x\) axis is mirrored around that axis to achieve positive value. The positive line segment of function \(g(x)\) is not mirrored since the absolute value of positive real number is the same.
• Example 2 Draw the graphs of two linear functions with absolute values i.e. \(f(x) = |3x|\) and \(g(x) = |3x| + 5\)
  Solution To represent these functions graphically the values of these functions must be determined for different values of variable \(x\). The values of function \(f(x)\) obtained for different values of \(x\) in range from -10 to 10. \begin{eqnarray} x = -10 \Rightarrow f(-10) &=& |3\cdot -10| = |-30| = 30\\ x = -8 \Rightarrow f(-8) &=& |3 \cdot (-8)| = |-24| = 24 \\ x = -6 \Rightarrow f(-6) &=& |3\cdot (-6)|= |-18| = 18 \\ x = -4 \Rightarrow f(-4) &=& |3\cdot(-4)| = |-12| = 12 \\ x = -2 \Rightarrow f(-2) &=& |3\cdot (-2)| = |-6| = 6 \\ x= 0 \Rightarrow f(0) &=& |3\cdot 0| = 0 \\x =2 \Rightarrow f(2) &=& |3\cdot 2 |= 6 \\x = 4 \Rightarrow f(4) &=& |3\cdot 4| = 12 \\ x = 6 \Rightarrow f(6) &=& |3\cdot 6| = 18 \\ x = 8 \Rightarrow f(8) &=& |3\cdot 8 |= 24 \\ x = 10 \Rightarrow f(10) &=& |3\cdot 10|= 30\end{eqnarray} The values of function \(g(x)\) obtained for different values of \(g(x)\) obtaied for different values of \(x\) in range from -10 to 10. \begin{eqnarray} x = -10 \Rightarrow f(-10) &=& |3\cdot -10| + 5 = |-30| + 5 = 35\\ x = -8 \Rightarrow g(-8) &=& |3 \cdot (-8)| +5 = |-24|+5 = 29 \\ x = -6 \Rightarrow g(-6) &=& |3\cdot (-6)| + 5= |-18| +5 = 23 \\ x = -4 \Rightarrow g(-4) &=& |3\cdot(-4)| + 5 = |-12| + 5 = 17 \\ x = -2 \Rightarrow g(-2) &=& |3\cdot (-2)| + 5 = |-6| + 5 = 11 \\ x= 0 \Rightarrow f(0) &=& |3\cdot 0|+5 = 5 \\x =2 \Rightarrow g(2) &=& |3\cdot 2 | + 5 = 11 \\x = 4 \Rightarrow g(4) &=& |3\cdot 4| +5 = 17 \\ x = 6 \Rightarrow g(6) &=& |3\cdot 6|+5 = 23 \\ x = 8 \Rightarrow g(8) &=& |3\cdot 8 | + 5= 29 \\ x = 10 \Rightarrow g(10) &=& |3\cdot 10| + 5 = 35\end{eqnarray} The calculated solutions for both functions are given in following table and the functions are graphically represented in following figure.

  \(x\)	\(f(x)\)	\(g(x)\)
  -10	30	35
  -8	24	29
  -6	18	23
  -4	12	17
  -2	6	11
  0	0	5
  2	6	11
  4	12	17
  6	18	23
  8	24	29
  10	30	35

  
  
  
  From calculated values, table and figure it can be seen that the values of \(g\) can be obtained by increasing the values of the function \(f(x)\) by 5. This means that we obtain the graph of the function \(g(x)\) by raising the graph \(f(x)\) along the y-axis by the value 5.
• Example 3 Draw the graph of the function \(f(x) = -|x| -5\)
  Solution The procedure is the same as in previous example. Fitst the values of the function \(f(x)\) will be obtained for differennt values of \(x\). Then the calculated reuslts willl be represented graphically and will be used to draw a graph of the function. However, instead of directly drawing the function \(f(x)\) we will start with the function \(h(x) = -|x|\). The values of the function \(h(x)\) are calculated for different values of \(x\) in range from -10 to 10. \begin{eqnarray} x = -10 \Rightarrow h(-10) &=& -|-10| = -10\\ x = -8 \Rightarrow h(-8) &=& -|-8| = -8 \\ x = -6 \Rightarrow h(-6) &=& -|-6| = - 6\\ x = -4 \Rightarrow h(-4) &=& -|-4| = - 4\\ x = -2 \Rightarrow h(-2) &=& -|-2| = -2\\ x = 0 \Rightarrow h(0) &=& -|0| = 0\\ x = 2 \Rightarrow h(2) &=& -|2| = -2 \\ x = 4 \Rightarrow h(4) &=& -|4| = - 4\\ x = 6 \Rightarrow h(6) &=& -|6| = -6\\ x = 8 \Rightarrow h(8) &=& -|8| = -8\\ x = 10\Rightarrow h(10) &=& -|10| = -10 \end{eqnarray} The values of the function \(f(x)\) are claculated for different values of \(x\) in range from -10 to 10. \begin{eqnarray} x = -10 \Rightarrow f(-10) &=& -|-10| + 5 = -5\\ x = -8 \Rightarrow f(-8) &=& -|-8| + 5 = -3 \\ x = -6 \Rightarrow f(-6) &=& -|-6| + 5 = - 1\\ x = -4 \Rightarrow f(-4) &=& -|-4| + 5 = 1\\ x = -2 \Rightarrow f(-2) &=& -|-2| + 5 = 3\\ x = 0 \Rightarrow f(0) &=& -|0| + 5 = 5\\ x = 2 \Rightarrow f(2) &=& -|2| + 5 = 3 \\ x = 4 \Rightarrow f(4) &=& -|4| + 5 = 1\\ x = 6 \Rightarrow f(6) &=& -|6| + 5 = -1\\ x = 8 \Rightarrow f(8) &=& -|8| + 5 = -3\\ x = 10\Rightarrow f(10) &=& -|10| + 5 = -5 \end{eqnarray} The calculated output values of both functions i.e \(h(x)\) and \(f(x)\) are given in following table while graphs of these functions is shown in following figure.

  \(x\)	\(h(x)\)	\(f(x)\)
  -10	-10	-5
  -8	-8	-3
  -6	-6	-1
  -4	-4	1
  -2	-2	3
  0	0	5
  2	-2	3
  4	-4	1
  6	-6	-1
  8	-8	-3
  10	-10	-5

  
  
  
  From obtained solutions it can be concluded that the graph of the function \(f(x) = -|x| + 5\) can be achieved by plotting the graph of the function \(h(x) = -|x|\) and translating it for 5 units upwards along the \(y\) axis.
• Example 4 Write the function \(f(x) = |x-3|\) without absolute value.
  Solution To solve the equation we have to assume that the expression inside absolute brackets is postivie and negative. \begin{eqnarray} x - 3 &<& 0 \Rightarrow |x-3| = -(x-3) = 3 - x \\ x-3 &\geq& 0 \Rightarrow |x-3| = x-3\\ f(x) &=& \begin{cases} 3-x, \quad \mathrm{for} \quad x < 3\\ x-3, \quad \mathrm{for} \quad x \geq 3 \end{cases} \end{eqnarray}
• Example 5 Write the function \(f(x) = |2x-5| + 2x -3 without absolute value brakets.
  Solution As in the previous case we have to investigate if the expression inside absolute brackets are positive or negative. \begin{eqnarray} 2x-5 &<&0 \Rightarrow |2x-5| = -(2x-5) = 5-2x\\ 2x-5 &\geq& 0 \Rightarrow |2x-5| = 2x-5\\ f(x)&=& \begin{cases} 2, \quad \mathrm{for} \quad 2x-5 < 0 \\ 4x-8 \quad \mathrm{for} \quad 2x-5 \geq 0 \end{cases} \end{eqnarray}
• Example 6 Draw the graph of the function \(f(x) = \sqrt{x^2 - 6x + 9}\)
  Solution By defintion the square root of the variable to the power of two is the absolute value. This deifinition can be written in mathematical form as $$ \sqrt{x^2} = |x|. $$ So, before writting the function with absolute value the term under the square root must be simplifed. \begin{eqnarray}x^2 - 6x + 9 &=& (x-3)^2 \end{eqnarray} The function can be written as: \begin{eqnarray} f(x) &=& \sqrt{x^2 - 6x + 9} \\ f(x) &=& \sqrt{(x-3)^2} \\ f(x) &=& |x-3| \end{eqnarray} Before we draw the graph of the function the output values of the functions must be calculated for different values of argument \(x\). The output values of the function for five arbitrary chosne values of \(x\) are: \begin{eqnarray} x&=& 0 \Rightarrow f(0) = |0-3| = 3 \\ x &=& 3 \Rightarrow f(3) = |3-3| = 0 \\ x &=& 6 \Rightarrow f(6) = |6-3| = 3 \end{eqnarray} The graph of the function is shown in following figure.
  
  
  
  
• Example 7 From following figure determine which is graph of the function \(f(x) = |3x| - 2\)
  
  
  
  
  Solution Since the function we are looking for has the -2 value for the argument value of 0 the function can easily be distinguished from the figure (The orange one). However, the best way to find the funciton is to choose some arbitrary values of argument \(x\) to calculate the output \(f(x)\). Here we have chosen -1, 0 and 1 as the values of \(x\). \begin{eqnarray} x &=& -1 \Rightarrow f(-1) = |3\cdot (-1)|-2 = 1 \\ x &=& 0 \Rightarrow f(0) = |3\cdot 0| -2 = -2 \\ x &=& 1 \Rightarrow f(1) = |3\cdot 1| -2 = 1 \end{eqnarray} From the obtained values it can easily be concluded that the graph of the function \(f(x) = |3x|-2 \) is represented with the orange lines.
• Example 8 Determine the vertex of the function \(f(x) = |x+4| +2 \)
  Solution \begin{eqnarray} f(-6) &=& |-6+4| + 2 = 4\\ f(-5) &=& |-5+4| +2 = 3 \\f(-4) &=& |-4+4| +2 = 2 \\ f(-3) &=& |-3+4| +2 = 3\\ f(-2) &=& |-2+4| + 2 = 4 \\f(0) &=& |0 +4|+2 = 6 \end{eqnarray} The sequence of function outputs for different values of argument \(x\) shows that the minimum value achieved by function \(f(x)\) is 2 with \(x\) value of -4. So the vertex of the function \(f(x) = |x+4| +2 \) is \(V(-4,2)\).
• Example 9 Draw the graph of the function \(f(x) = \sqrt{25+10x + x^2}\)
  Solution To draw the graph of the function the function outputs must be calculated for arbitrary values of \(x\). But before that the function will be simplified. \begin{eqnarray}f(x) &=& \sqrt{25 + 10x + x^2}\\ f(x) &=& \sqrt{(5+x)^2} \\ f(x) &=& |5+x| \end{eqnarray} Now the outputs are calculated for arbitrary chosen \(x\) values i.e. -10, -5, 0, 5 and 10. \begin{eqnarray} x&=& -10 \Rightarrow f(-10) = |5-10| = 5 \\ x&=& -5 \Rightarrow f(-5) = |5-5| = 0\\ x&=& 0 \Rightarrow f(0) = |5-0| = 5 \\ x&=& 5 \Rightarrow f(5) = |5+5| = 10 \\ x&=& 10 \Rightarrow f(10) = |5+10| = 15 \end{eqnarray} Now that the outputs are obtained for different values of \(x\) the graph of the function is shown in following figure.
  
  
  
  
• Example 10 Determine the unkown coordinate of the point A(-2,y) which is the part of the function \(f(x) = \left|\frac{2}{3}x - 1\right|\).
  Solution The procedure in this example very simple. All we have to do is insert the value of \(x=-2\) in the function and calculate the output. \begin{eqnarray} x = -2 \Rightarrow f(-2) &=& \left|\frac{2}{3}(-2) - 1\right|\\ f(-2) &=& \left|\frac{-4}{3} -1 \right|\\ f(-2) &=& \left|\frac{-4-3}{3}\right|\\ f(-2) &=& \left|\frac{-7}{3}\right| = \frac{7}{3}\\ &A&\left(-2, \frac{7}{3}\right) \end{eqnarray}