1.6: Absolute Value Functions
- Foundational Topics
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-
Arithmetic
- Percents
-
Arithmetic
In this section you will:
- Graph an absolute value function.
- Solve an absolute value equation.
- Solve an absolute value inequality.
Figure \( \PageIndex{ 1 } \) Distances in deep space can be measured in all directions. As such, it is useful to consider distance in terms of absolute values. (credit: "s58y"/Flickr)
Until the 1920s, the so-called spiral nebulae were believed to be clouds of dust and gas in our own galaxy, some tens of thousands of light years away. Then, astronomer Edwin Hubble proved that these objects are galaxies in their own right, at distances of millions of light years. Today, astronomers can detect galaxies that are billions of light years away. Distances in the universe can be measured in all directions. As such, it is useful to consider distance as an absolute value function. In this section, we will investigate absolute value functions.
Focusing on Calculus - Understanding Absolute Value
Recall that in its basic form \(f(x)=| x |\), the absolute value function, is one of our toolkit functions. The absolute value function is commonly thought of as providing the distance the number is from zero on a number line. Algebraically, for whatever the input value is, the output is the value without regard to sign.
The
absolute value function
can be defined as a piecewise function\[f(x)=| x |= \begin{cases}
x, & \text{ if } x \geq 0 \\[6pt]
−x, & \text{ if } x<0 \\[6pt]
\end{cases} \nonumber \]
Describe all values \(x\) within or including a distance of 4 from the number 5.
- Solution
-
We want the distance between \(x\) and 5 to be less than or equal to 4. We can draw a number line, such as the one in Figure \( \PageIndex{ 2 } \), to represent the condition to be satisfied.
Figure \( \PageIndex{ 2 } \)The distance from \(x\) to 5 can be represented using the absolute value as \(| x−5 |.\) We want the values of \(x\) that satisfy the condition \(| x−5 | \leq 4.\)
Note that\[ \begin{array}{rclcrcl}
−4 & \leq & x−5 & \quad & x−5 & \leq & 4 \\[6pt]
1 & \leq & x & \quad & x & \leq & 9. \\[6pt]
\end{array} \nonumber \]So \(| x−5 | \leq 4\) is equivalent to \(1 \leq x \leq 9\).
However, mathematicians generally prefer absolute value notation.
Describe all values \(x\) within a distance of 3 from the number 2.
Electrical parts, such as resistors and capacitors, come with specified values of their operating parameters: resistance, capacitance, etc. However, due to imprecision in manufacturing, the actual values of these parameters vary somewhat from piece to piece, even when they are supposed to be the same. The best that manufacturers can do is to try to guarantee that the variations will stay within a specified range, often \(\pm 1\%\), \(\pm 5\%\), or \(\pm 10\%\).
Suppose we have a resistor rated at 680 ohms, \(\pm 5\%\). Use the absolute value function to express the range of possible values of the actual resistance.
- Solution
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5% of 680 ohms is 34 ohms. The absolute value of the difference between the actual and nominal resistance should not exceed the stated variability, so, with the resistance \(R\) in ohms,\[| R−680 | \leq 34. \nonumber \]
Students who score within 20 points of 80 will pass a test. Write this as a distance from 80 using absolute value notation.
Graphing an Absolute Value Function
The most significant feature of the absolute value graph is the corner point at which the graph changes direction. This point is shown at the origin in Figure \( \PageIndex{ 3 } \).
Figure \( \PageIndex{ 3 } \)
Figure \( \PageIndex{ 4 } \) shows the graph of \(y=2| x–3 |+4.\) The graph of \(y=| x |\) has been shifted right 3 units, vertically stretched by a factor of 2, and shifted up 4 units. This means that the corner point is located at \(( 3,4 )\) for this transformed function.
Figure \( \PageIndex{ 4 } \)
Write an equation for the function graphed in Figure \( \PageIndex{ 5 } \).
Figure \( \PageIndex{ 5 } \)
- Solution
-
The basic absolute value function changes direction at the origin, so this graph has been shifted to the right 3 units and down 2 units from the basic toolkit function. See Figure \( \PageIndex{ 6 } \).
Figure \( \PageIndex{ 6 } \)We also notice that the graph appears vertically stretched, because the width of the final graph on a horizontal line is not equal to 2 times the vertical distance from the corner to this line, as it would be for an unstretched absolute value function. Instead, the width is equal to 1 times the vertical distance as shown in Figure \( \PageIndex{ 7 } \).
Figure \( \PageIndex{ 7 } \)From this information we can write the equation as \(f(x)=2|x−3|−2\), treating the stretch as a vertical stretch.
In Example \( \PageIndex{ 3 } \), we could have opted to treat the stretch as a horizontal compression. In that case, the function would have been \( f(x)=|2(x−3)|−2 \). Note that this equation is algebraically equivalent to what we arrived at in our solution. This result, that the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression, is not a rule for all functions - it just happens to work for absolute value (and linear) functions. Note also that if the vertical stretch factor is negative, there is also a reflection about the x-axis.
If we couldn’t observe the stretch of the function from the graphs, could we algebraically determine it?
Yes. If we are unable to determine the stretch based on the width of the graph, we can solve for the stretch factor by putting in a known pair of values for \(x\) and \(f(x)\).\[f(x)=a|x−3|−2. \nonumber \]Now substituting in the point \( (1, 2) \), we get\[ \begin{array}{rrclcl}
& 2 & = & a| 1−3 |−2 & \quad & \left( \text{substituting} \right) \\[6pt]
\implies & 4 & = & 2a & \quad & \left( \text{simplifying and adding }2 \text{ to both sides} \right) \\[6pt]
\implies & 2 & = & a & \quad & \left( \text{dividing both sides by }2 \right) \\[6pt]
\end{array} \nonumber \]
Write the equation for the absolute value function that is horizontally shifted left 2 units, is vertically flipped, and vertically shifted up 3 units.
Do the graphs of absolute value functions always intersect the vertical axis? The horizontal axis?
Yes, they always intersect the vertical axis. The graph of an absolute value function will intersect the vertical axis when the input is zero.
No, they do not always intersect the horizontal axis. The graph may or may not intersect the horizontal axis, depending on how the graph has been shifted and reflected. It is possible for the absolute value function to intersect the horizontal axis at zero, one, or two points (see Figure \( \PageIndex{ 8 } \)).
Figure \( \PageIndex{ 8(a) } \): The absolute value function does not intersect the horizontal axis.
Figure \( \PageIndex{ 8(b) } \): The absolute value function intersects the horizontal axis at one point.
Figure \( \PageIndex{ 8(c) } \): The absolute value function intersects the horizontal axis at two points.
Is the absolute value function one-to-one?
Solving an Absolute Value Equation
Now that we can graph an absolute value function, we will learn how to solve an absolute value equation. To solve an equation such as \(8=| 2x−6 |\), we notice that the absolute value will be equal to 8 if the quantity inside the absolute value is 8 or -8. This leads to two different equations we can solve independently.\[ \begin{array}{rrclcrcl}
& 2x−6 & = & 8 & \text{ or } & 2x−6 & = & −8 \\[6pt]
\implies & 2x & = & 14 & & 2x & = & −2 \\[6pt]
\implies & x &= &7 & & x & = & −1 \\[6pt]
\end{array} \nonumber \]Knowing how to solve problems involving absolute value functions is useful. For example, we may need to identify numbers or points on a line that are at a specified distance from a given reference point.
An
absolute value equation
is an equation in which the unknown variable appears in absolute value bars. For example,\[ \begin{array}{rcl}
| x | & = & 4,\\[6pt]
| 2x−1 | & = & 3, \text{ and}\\[6pt]
| 5x+2 |−4 & = & 9. \\[6pt]
\end{array} \nonumber \]To solve an absolute value equation, we rely on the function definition of the absolute value. Namely,\[=| \blacksquare |= \begin{cases}
\blacksquare, & \text{ if } \blacksquare \geq 0 \\[6pt]
−\blacksquare, & \text{ if } \blacksquare<0 \\[6pt]
\end{cases} \nonumber \]I specifically chose to use \( \blacksquare \) in this version of the definition instead of \( x \) because most students new to the concept of the piecewise definition of the absolute value tend to think that the conditions must always be \( x \geq 0 \) and \( x < 0 \) -
this is not true
.
Let's illustrate the process through an example.
For the function \(f(x)=| 4x+1 |−7\), find the values of \(x\) such that \(f(x)=0\).
- Solution
-
We set \( f(x) = 0 \) and solve.\[ \begin{array}{rrclcl}
& 0 & = & |4x + 1| - 7 & \quad & \left( \text{setting }f(x) = 0 \right) \\[6pt]
\implies & 7 & = & |4x + 1| & \quad & \left( \text{adding }7 \text{ to both sides} \right) \\[6pt]
\end{array} \nonumber \]Now that we have the absolute value isolated, let's recall the piecewise function definition of the absolute value.\[| \blacksquare |= \begin{cases}
\blacksquare, & \text{ if } \blacksquare \geq 0 \\[6pt]
−\blacksquare, & \text{ if } \blacksquare<0 \\[6pt]
\end{cases} \nonumber \]In our case, \( \blacksquare = 4x + 1 \). Hence,\[| 4x + 1 |= \begin{cases}
4x + 1, & \text{ if } 4x + 1 \geq 0 \\[6pt]
−(4x + 1), & \text{ if } 4x + 1<0 \\[6pt]
\end{cases} \nonumber \]That is,\[| 4x + 1 |= \begin{cases}
4x + 1, & \text{ if } x \geq -\frac{1}{4} \\[6pt]
−(4x + 1), & \text{ if } x < -\frac{1}{4} \\[6pt]
\end{cases} \nonumber \]This tells us that our absolute value equation splits into two possibilies:\[ \begin{array}{ccc}
\text{For }x \geq - \frac{1}{4} & \quad & \text{For }x < -\frac{1}{4} \\[6pt]
\begin{array}{rrcl}
& 7 & = & |4x + 1| \\[6pt]
\implies & 7 & = & 4x + 1 \\[6pt]
\implies & 6 & = & 4x \\[6pt]
\implies & \frac{3}{2} & = & x \\[6pt]
& & & \\[6pt]
\end{array} & \quad & \begin{array}{rrcl}
& 7 & = & |4x + 1| \\[6pt]
\implies & 7 & = & -(4x + 1) \\[6pt]
\implies & -7 & = & 4x + 1 \\[6pt]
\implies & -8 & = & 4x \\[6pt]
\implies & -2 & = & x \\[6pt]
\end{array} \\[6pt]
\end{array} \nonumber \]The function outputs 0 when \(x=1.5\) or \(x=−2.\) See Figure \( \PageIndex{ 9 } \).
Figure \( \PageIndex{ 9 } \)
For the function \(f(x)=| 2x−1 |−3\), find the values of \(x\) such that \(f(x)=0.\)
Should we always expect two answers when solving \(| A |=B?\)
No. We may find one, two, or even no answers. For example, there is no solution to \(2+| 3x−5 |=1.\)
Solve \(1=4| x−2 |+2\).
- Solution
-
Isolating the absolute value on one side of the equation gives the following.\[ \begin{array}{rrcl}
& 1 & = & 4| x−2 |+2 \\[6pt]
\implies & −1 & = & 4| x−2 | \\[6pt]
\implies & −\dfrac{1}{4} & = & | x−2 | \\[6pt]
\end{array} \nonumber \]The absolute value always returns a positive value, so it is impossible for the absolute value to equal a negative value. At this point, we notice that this equation has no solutions.
In Example \( \PageIndex{ 5 } \), if \(f(x)=1\) and \(g(x)=4| x−2 |+2\) were graphed on the same set of axes, would the graphs intersect?
No. The graphs of \(f\) and \(g\) would not intersect, as shown in Figure \( \PageIndex{ 10 } \). This confirms, graphically, that the equation \(1=4| x−2 |+2\) has no solution.
Figure \( \PageIndex{ 10 } \)
Find where the graph of the function \(f(x)=−| x+2 |+3\) intersects the horizontal and vertical axes.
Focusing on Calculus - Solving an Absolute Value Inequality
Absolute value equations may not always involve equalities. Instead, we may need to solve an equation within a range of values. We would use an absolute value inequality to solve such an equation. An absolute value inequality is an equation of the form\[ |A|<B, \quad |A| \leq B, \quad |A|>B, \quad \text{or }|A| \geq B, \nonumber \]where an expression \(A\) (and possibly, but not usually, \(B\) ) depends on a variable \(x\). Solving the inequality means finding the set of all \(x\) that satisfy the inequality. Usually this set will be an interval or the union of two intervals.
There are two basic approaches to solving absolute value inequalities: graphical and algebraic. The advantage of the graphical approach is we can read the solution by interpreting the graphs of two functions. The disadvantage of this approach is that, when graphing by hand, the solutions are likely not going to be accurate. The advantage of the algebraic approach is it yields solutions that may be difficult to read from the graph.
For example, we know that all numbers within 200 units of 0 may be expressed as\[| x |<200 \text{ or } −200<x<200. \nonumber \]Suppose we want to know all possible returns on an investment if we could earn some amount of money within $200 of $600. We can solve algebraically for the set of values \(x\) such that the distance between \(x\) and 600 is less than 200. We represent the distance between \(x\) and 600 as \(| x−600 |\). Thus, we want \(|x−600|<200\), or, using inequality notation, \( -200<x−600<200\). Solving this inequality, we get\[ \begin{array}{rrccclcl}
& -200 & < & x - 600 & < & 200 & & \\[6pt]
\implies & 400 & < & x & < & 800 & \quad & \left( \text{adding }600 \text{ to all three "sides"} \right) \\[6pt]
\end{array} \nonumber \]This means our returns would be between $400 and $800.
Sometimes an absolute value inequality problem will be presented to us in terms of a shifted and/or stretched or compressed absolute value function, where we must determine for which values of the input the function’s output will be negative or positive.
Solve \(|x−5|<4\).
- Solution
-
Our approach will duplicate our work in Example \( \PageIndex{ 5 } \). Since the inequality is already isolated, we immediately dive into the piecewise definition of the absolute value:\[| \blacksquare |= \begin{cases}
\blacksquare, & \text{ if } \blacksquare \geq 0 \\[6pt]
−\blacksquare, & \text{ if } \blacksquare<0 \\[6pt]
\end{cases} \nonumber \]In our case, \( \blacksquare = x - 5 \). Hence,\[| x - 5 |= \begin{cases}
x - 5, & \text{ if } x - 5 \geq 0 \implies x \geq 5 \\[6pt]
−(x - 5), & \text{ if } x - 5 < 0 \implies x < 5 \\[6pt]
\end{cases} \nonumber \]This tells us that our absolute value inequality splits into two possibilies:\[ \begin{array}{ccc}
\text{For }x \geq 5 & \quad & \text{For }x < 5 \\[6pt]
\begin{array}{rrcl}
& |x - 5| & < & 4 \\[6pt]
\implies & x - 5 & < & 4 \\[6pt]
\implies & x & < & 9 \\[6pt]
& & & \\[6pt]
\end{array} & \quad & \begin{array}{rrcl}
& |x - 5| & < & 4 \\[6pt]
\implies & -(x - 5) & < & 4 \\[6pt]
\implies & x - 5 & > & -4 \\[6pt]
\implies & x & > & 1 \\[6pt]
\end{array} \\[6pt]
\end{array} \nonumber \]To summarize, we are okay with \( x \geq 5 \) as long as \( x < 9 \). Additionally, we are okay with \( x < 5 \) as long as \( x > 1 \). Combining this information, our solution interval is \( \left( 1 , 9 \right) \).
The approach we used in our solution to Example \( \PageIndex{ 6 } \) is what I consider to be a "preferred method." This is because it reinforces the piecewise definition of the absolute value (which is something you must know to be successful in Calculus); however, it is best to show you another approach used by many other instructors and students.
First, we will need to know where the corresponding equality is true. That is, we need to solve \( |x - 5| = 4 \). The process is almost identical to what we did in Example \( \PageIndex{ 6 } \).\[| x - 5 |= \begin{cases}
x - 5, & \text{ if } x - 5 \geq 0 \implies x \geq 5 \\[6pt]
−(x - 5), & \text{ if } x - 5 < 0 \implies x < 5 \\[6pt]
\end{cases} \nonumber \]This tells us that our absolute value equation splits into two possibilies:\[ \begin{array}{ccc}
\text{For }x \geq 5 & \quad & \text{For }x < 5 \\[6pt]
\begin{array}{rrcl}
& |x - 5| & = & 4 \\[6pt]
\implies & x - 5 & = & 4 \\[6pt]
\implies & x & = & 9 \\[6pt]
& & & \\[6pt]
\end{array} & \quad & \begin{array}{rrcl}
& |x - 5| & = & 4 \\[6pt]
\implies & -(x - 5) & = & 4 \\[6pt]
\implies & x - 5 & = & -4 \\[6pt]
\implies & x & = & 1 \\[6pt]
\end{array} \\[6pt]
\end{array} \nonumber \]After determining that the absolute value is equal to 4 at \(x=1\) and \(x=9\), we know the graph can change only from being less than 4 to greater than 4 at these values. This divides the number line up into three intervals:\[x<1, \quad 1<x<9, \quad \text{and } x>9. \nonumber \]To determine when the function is less than 4, we could choose a value in each interval and see if the output is less than or greater than 4, as shown in Table \( \PageIndex{ 1 } \).
| Interval test \(x\) | Test value for \(x\) | \(<4\) or \(>4\)? | |
|---|---|---|---|
| \(x<1\) | 0 | \(| 0−5 |=5\) | Greater than |
| \(1<x<9\) | 6 | \(| 6−5 |=1\) | Less than |
| \(x>9\) | 11 | \(| 11−5 |=6\) | Greater than |
Because \(1<x<9\) is the only interval in which the output at the test value is less than 4, we can conclude that the solution to \(| x−5 |<4\) is \(1<x<9\), or \(( 1,9 )\).
To use a graph, we can sketch the function \(f(x)=| x−5 |\). To help us see where the outputs are 4, the line \(g(x)=4\) could also be sketched as in Figure \( \PageIndex{ 11 } \).
Figure \( \PageIndex{ 11 } \): Graph to find the points satisfying an absolute value inequality.
We can see the following:
- The output values of the absolute value are equal to 4 at \(x=1\) and \(x=9\).
- The graph of \(f\) is below the graph of \(g\) on \(1<x<9\). This means the output values of \(f(x)\) are less than the output values of \(g(x)\).
- The absolute value is less than or equal to 4 between these two points, when \(1<x<9\). In interval notation, this would be the interval \(( 1,9 )\).
For absolute value inequalities,\[ \begin{array}{rrccclcrcccl}
& & & |x−A| & < & C & \quad & |x−A| & > & C & & \\[6pt]
\implies & −C & < & x−A & < & C & \quad & x−A & < & −C & \text{or} & x−A > C \\[6pt]
\end{array} \nonumber \]The \(<\) or \(>\) symbol may be replaced by \(\leq\) or \( \geq \).
So, for Example \( \PageIndex{ 6 } \), we could use this alternative approach.\[ \begin{array}{rrccclcl}
& & & |x−5| & < & 4 & \quad & \\[6pt]
\implies & −4 & < & x−5 & < & 4 & \quad & \left(\text{rewriting by removing the absolute value bars}\right) \\[6pt]
\implies & −4+5 & < & x−5+5 & < & 4+5 & \quad & \left(\text{adding }5\text{ to all three "sides"}\right) \\[6pt]
\implies & 1 & < & x & < & 9 & & \\[6pt]
\end{array} \nonumber \]
Solve \(| x+2 | \leq 6\).
Given the function \(f(x)= -\frac{1}{2} | 4x−5 |+3\), determine the \(x\)-values for which the function values are negative.
- Solution
-
We are trying to determine where \(f(x)<0\), which is when \( -\frac{1}{2} |4x−5|+3<0\). We begin by isolating the absolute value.\[ \begin{array}{rrclcl}
& -\dfrac{1}{2} |4x−5| & < & −3 & \quad & \left(\text{subtracting }3\text{ from both sides}\right)\\[6pt]
\implies & |4x−5| & > & 6 & \quad & \left( \text{multiplying both sides by }-2\text{ and switching the inequality sign} \right)\\[6pt]
\end{array} \nonumber \]Using the definition of the absolute value, we get\[| 4x - 5 |= \begin{cases}
4x - 5, & \text{ if } 4x - 5 \geq 0 \implies x \geq \frac{5}{4} \\[6pt]
−(4x - 5), & \text{ if } 4x - 5 < 0 \implies x < \frac{5}{4} \\[6pt]
\end{cases} \nonumber \]Thus,\[ \begin{array}{ccc}
\text{For }x \geq \frac{5}{4} & \quad & \text{For }x < \frac{5}{4} \\[6pt]
\begin{array}{rrcl}
& |4x - 5| & > & 6 \\[6pt]
\implies & 4x - 5 & > & 6 \\[6pt]
\implies & 4x & > & 11 \\[6pt]
\implies & x & > & \dfrac{11}{4} \\[6pt]
& & & \\[6pt]
& & & \\[6pt]
\end{array} & \quad & \begin{array}{rrcl}
& |4x - 5| & > & 6 \\[6pt]
\implies & -(4x - 5) & > & 6 \\[6pt]
\implies & 4x - 5 & < & -6 \\[6pt]
\implies & 4x & < & -1 \\[6pt]
\implies & x & < & -\dfrac{1}{4} \\[6pt]
& & & \\[6pt]
\end{array} \\[6pt]
\end{array} \nonumber \]The leftmost solution tells us that as long as \( x \geq \frac{5}{4} \), the original inequality is satisfied for \( x > \frac{11}{4} \). The rightmost solution tells us that as long as \( x < \frac{5}{4} \), the original inequality is satisfied for \( x < -\frac{1}{4} \). Thus, our solution is \( \left( -\infty, -\frac{1}{4} \right) \cup \left( \frac{11}{4}, \infty \right) \).Had we decided to using the graphing method, we can examine the graph of \(f\) to observe where the output is negative. We will observe where the branches are below the \( x \)-axis. Notice that it is not even important exactly what the graph looks like, as long as we know that it crosses the horizontal axis at \(x= -\frac{1}{4} \) and \(x = \frac{11}{4} \) and that the graph has been reflected vertically. See Figure \( \PageIndex{ 12 } \) .
Figure \( \PageIndex{ 12 } \)We observe that the graph of the function is below the \( x \)-axis left of \(x = -\frac{1}{4} \) and right of \(x = \frac{11}{4} \). This means the function values are negative to the left of the first horizontal intercept at \(x = -\frac{1}{4} \), and negative to the right of the second intercept at \(x = \frac{11}{4} \). This gives us the solution to the inequality.\[ x < -\dfrac{1}{4} \text{ or } x > \dfrac{11}{4}. \nonumber \]In interval notation, this would be \( \left( -\infty, -\frac{1}{4} \right) \cup \left( \frac{11}{4}, \infty \right) \).
Solve \(−2| k−4 | \leq −6\).
Focusing on Calculus - Revisiting the Difference Quotient
Find, and simplify, the difference quotient for \(f(x)= -\frac{1}{2} | 4x−5 |+3\) assuming that \( x < \frac{5}{4} \) and \( h < 0 \).
- Solution
-
\[ \begin{array}{rclcl}
\dfrac{f(x + h) - f(x)}{h} & = & \dfrac{\left(-\frac{1}{2} |4(x + h) - 5| + 3\right) - \left( -\frac{1}{2} | 4x−5 |+3 \right)}{h} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{-\frac{1}{2} |4(x + h) - 5| + 3 + \frac{1}{2} | 4x−5 | - 3 }{h} & \quad & \left( \text{distributing} \right) \\[6pt]
& = & \dfrac{-\frac{1}{2} |4(x + h) - 5| + \frac{1}{2} | 4x−5 |}{h} & \quad & \left( \text{simplifying} \right) \\[6pt]
& = & \dfrac{2}{2} \cdot \dfrac{-\frac{1}{2} |4(x + h) - 5| + \frac{1}{2} | 4x−5 |}{h} & \quad & \left( \text{simplifying the compound fraction} \right) \\[6pt]
& = & \dfrac{-|4(x + h) - 5| + | 4x−5 |}{2h} & \quad & \left( \text{distributing} \right) \\[6pt]
& = & \dfrac{-|4x + 4h - 5| + | 4x−5 |}{2h} & \quad & \left( \text{distributing} \right) \\[6pt]
\end{array} \nonumber \]At this point, we are stuck and need to refer to the piecewise definition of the absolute value function.\[| 4x - 5 |= \begin{cases}
4x - 5, & \text{ if } 4x - 5 \geq 0 \implies x \geq \frac{5}{4} \\[6pt]
−(4x - 5), & \text{ if } 4x - 5 < 0 \implies x < \frac{5}{4} \\[6pt]
\end{cases} \nonumber \]Since we were told \( x < \frac{5}{4} \), \( |4x - 5| = -(4x - 5) = -4x + 5 \). Turning our attention to the other absolute value, we get the following:\[| 4x + 4h - 5 |= \begin{cases}
4x + 4h - 5, & \text{ if } 4x + 4h - 5 \geq 0 \implies 4x \geq 5 - 4h \implies x \geq \frac{5}{4} - h \\[6pt]
−(4x + 4h - 5), & \text{ if } 4x + 4h - 5 < 0 \implies 4x < 5 - 4h \implies x < \frac{5}{4} - h \\[6pt]
\end{cases} \nonumber \]Since we are told that \( x < \frac{5}{4} \) and \( h < 0 \), it must be true that \( x + h < \frac{5}{4} \). Hence, \( x < \frac{5}{4} - h \) and so \( |4x + 4h - 5| = -(4x + 4h -5) = -4x - 4h + 5 \). Combining these two results, we get the following:\[ \begin{array}{rclcl}
\dfrac{f(x + h) - f(x)}{h} & = & \dfrac{-|4x + 4h - 5| + | 4x−5 |}{2h} & & \\[6pt]
& = & \dfrac{-\left(-4x - 4h + 5\right) + \left(-4x + 5\right)}{2h} & \quad & \left( \text{substituting} \right) \\[6pt]
& = & \dfrac{4x + 4h - 5 - 4x + 5}{2h} & \quad & \left( \text{distributing} \right) \\[6pt]
& = & \dfrac{4h}{2h} & \quad & \left( \text{combining like terms} \right) \\[6pt]
& = & \dfrac{4 \cancelto{1}{h}}{2\cancelto{1}{h}} & \quad & \left( \text{canceling like factors} \right) \\[6pt]
& = & 2 & & \\[6pt]
\end{array} \nonumber \]
Find, and simplify, the difference quotient for \(f(x)= -\frac{1}{2} | 4x−5 |+3\) assuming that \( x > \frac{5}{4} \) and \( h > 0 \).
Access these online resources for additional instruction and practice with absolute value.