The absolute value function is explored through definitions, examples and exercises with solutions at the bottom of the page. The graphing of absolute value functions |f(x)| is discussed through examples.

The absolute value of x written as |x| is the distance from zero to x. Since |x| gives a distance, it is always positive or equal to zero. Hence

| -5 | = 5

| 5 | = 5

| 0 | = 0

|a - b| is the distance between points at a and b on the number line.

|5 - 15| = | - 10| = 10

definition

\[ |x| = \left\{ \begin{array}{ll} x & x\geq 0 \\ -x & x\lt 0 \\ \end{array} \right. \]

which can be used to write

\[ |x - a| = \left\{ \begin{array}{ll} x - a & x\geq a \\ - (x - a) = a - x & x\lt a \\ \end{array} \right. \]

\( \sqrt {(x - a)^2} = |x - a| \)

- \( | x | \ge 0 \)
- \( \sqrt{ x^2} = | x | \)
- \( | x | = | - x | \)
- \( | a b | = | a || b | \)
- \( \left | \dfrac{a}{b} \right | = \dfrac{|a|}{|b|} , b \ne 0\)

- \( | x | = k \iff x = k \text{ or } x = - k \)
- \( | x | \le k \iff - k \le x \le k \)
- \( | x | ? k \iff x \le - k \text{ or } x \ge k \)

\[ f(x) = a x^2 + b x + c \] The exploration is carried out by changing the parameters a, b and c included in \( f(x)\) above.

- click on the button above "draw" to start.

- Use the sliders to set parameter a to zero, parameter b to zero and parameter c to a positive value; \( f(x)\) is a constant function. Compare the graph of f(x) in blue and that of \( h(x)= |f(x)| \) in red. Change c to a negative value and compare the graphs again. Use the definition of the absolute value functions to explain how can the graph of \( |f(x)| \) be obtained from the graph of \( f(x)\).

- Keep the value of a equal to zero, select non zero values for b to obtain a linear function . How can the graph of \( h(x) = |f(x)|\) be obtained from that of \( f(x)\)?

Hint: use the definition of the absolute value functions and reflection of a graph on the x-axis.

- Set b and c to zero and select a positive value for a to obtain a quadratic function . Why are the two graphs the same? (Hint: use the definition of the absolute value functions).

- Set b and c to zero and select a negative value for a to obtain a quadratic function . Why are the two graphs reflection of each other? (Hint: use the definition of the absolute value functions and reflection of a graph on the x-axis).

- Keep the values of a and b as in 5 above and change gradually c from zero to some positive values. How can the graph of \( h(x)=|f(x)|\) be obtained from that of \( f(x)\)?

- Select different values for a, b and c and explore.

Evaluate the expressions

- | - 9 |

- | 9 |

- | 2 - 6 |

- \( |- \dfrac{5}{2}| \)

- \( | 11 - 4^2 | \)

- \( 2^{| 2 - 7|} \)

- \( |2-19| - |9 - 20| \)

Substitute and Evaluate the expressions

- | x - 10 | + | 2 x | , x = - 7

- | - x | , x = 8

- \( \left |- \dfrac{|x - 2|}{|-x + 9|} \right | \) , x = -20

- \( | - 3 x - x^2 | \) , x = -3

- \( x^{| x - 7|} \) , x = - 2

Simplify the expressions

- | a - b | for a = b

- | a - b | for a > b

- | a - b | for b > a
- \( \sqrt {3^2} \)
- \( \sqrt {(- 3)^2} \)
- \( \sqrt { (12 - 20)^2 } \)
- \( | x^2 | \)
- \( | (-x)^2 | \)
- \( | (x + 1)^2 | \)
- \( \sqrt{(x - y + z)^2} \)
- \( \sqrt{sin^2(x)} \)

Solve the Equations

- \( |x - 2| = 4 \)
- \( |x + 5| = 0 \)
- \( |10 - x| = - 5 \)

Solve the Inequalities

- \( |x - 7| \gt 4 \)
- \( |x + 5| \lt 9 \)
- \( |x + 8| \lt - 5 \)
- \( |x + 8| \gt - 2 \)

Sketch the following pairs of functions in the same system of axes and explains the similarities and differences of the two graphs.

- \( f(x) = x - 1 , h(x) = |f(x)| \)

- \( f(x) = x^2 - 4 , h(x) = |f(x)| \)

- \( f(x) = \sin(x) , h(x) = |\sin(x)| \)

__Solutions to the Above Exercises__

Part 1

Evaluate the expressions

- | - 9 | = 9

- | 9 | = 9

- | 2 - 6 | = 4

- \( |- \dfrac{5}{2}| = \dfrac{5}{2} \)

- \( | 11 - 4^2 | = 5 \)

- \( 2^{| 2 - 7|} = 32 \)

- \( |2-19| - |9 - 20| = 6 \)

Substitute and Evaluate the expressions

- For x = - 7, | x - 10 | + | 2 x | = 31

- For x = 8 , | - x | = 8

- For x = - 20 , \( \left |- \dfrac{|x - 2|}{|-x + 9|} \right | = \dfrac{22}{29}\)

- For x = - 3 , \( | - 3 x - x^2 | = 0 \)

- For x = - 2 , \( x^{| x - 7|} = - 512\)

Simplify the expressions

- | a - b | = 0

- | a - b | = a - b

- | a - b | = b - a
- \( \sqrt {3^2} = |3| = 3 \)
- \( \sqrt {(- 3)^2} = |-3| = 3 \)
- \( \sqrt { (12 - 20)^2 = 8} \)
- \( | x^2 | = x^2\)
- \( | (-x)^2 | = x^2\)
- \( | (x + 1)^2 | = (x + 1)^2 \)
- \( \sqrt{(x - y + z)^2} = |x - y + z|\)
- \( \sqrt{sin^2(x)} = |sin(x)| \)

Solve the Equations

- \( |x - 2| = 4 \) , two solutions: x = - 2 and x = 6
- \( |x + 5| = 0 \) , one solution: x = - 5
- \( |10 - x| = - 5 \) , no real solutions.

Solve the Inequalities

- \( |x - 7| \gt 4 \) , solution set: x < 3 ? x > 11
- \( |x + 5| \lt 9 \) , solution set: -14 < x < 4
- \( |x + 8| \lt - 5 \) , solution set: {?}
- \( |x + 8| \gt - 2 \) , solution set: {all real numbers}

- \( f(x) = x - 1 , h(x) = |f(x)| = |x - 1| \)

For x - 1 ? 0 or x ? 1, h(x) = | x - 1 | = x - 1 and we therefore have h(x) = f(x) which we can be clearly seen from the graphs.

For x - 1 < 0 or x < 1 , h(x) = |x - 1| = - (x - 1) = - f(x).

Therefore for x < 1 h(x) = - f(x) which means that the graph of h is a reflection of the graph of f on the x axis and this is can be clearly seen on the graphs.

- \( f(x) = x^2 - 4 , h(x) = |f(x)| = |x^2 - 4| \)

For x^{2}- 4 ? 0 or -2 ? x ? 2 , h(x) = |(x^{2}- 4)| = - (x^{2}- 4) = - f(x) . Therefore, on the interval -2 ? x ? 2, the graph of h is a reflection of the graph of f. Outside this interval, the two graphs coincide.

- \( f(x) = \sin(x) , h(x) = |\sin(x)| \)

Over intervals -2? to -? , 0 to ? , 2? to 3?, ... sin(x) ? 0 and h(x) = sin(x) = f(x) therefore the graphs of h and f coincide.

For intervals - ? to 0 , ? to 2? ... sin(x) < 0 , h(x) = | sin(x) | = - sin(x) and the graph of h is a reflection of the graph of f on the x axis.

Absolute Value Equations And Inequalities.

Absolute Value.