permalinkWhether it's a washer, nut, bolt, or gear, when a machine part is made, it must be made to fit with all of the other parts of the system. Since no manufacturing process is perfect, there are small deviations from the norm when each piece is made. In fact, manufacturers have a range of acceptable values for each measurement of every screw, bolt, etc.
permalinkLet's say we were examining some new bolts just out of the factory. The manufacturer specifies that each bolt should be within a tolerance of 0.04 mm to 10 mm in diameter. So the lowest diameter that the bolt could be to make it through quality assurance is 0.04 mm smaller than 10 mm, which is 9.96 mm. Similarly, the largest diameter that the bolt could be is 0.04 mm larger than 10 mm, which is 10.04 mm.
permalinkTo write an equation that describes the minimum and maximum deviation from average, we want the difference between the actual diameter and the specification to be equal to 0.04 mm. Since absolute values are used to describe distances, we can summarize our thoughts mathematically as |x−10|=0.04, where x represents the diameter of an acceptably sized bolt, in millimeters. This equation says the same thing as the lowest diameter that the bolt could be to make it through quality assurance is 9.96 mm and the largest diameter that the bolt could be is 10.04 mm.
permalinkIn this section we will examine a variety of problems that relate to this sort of math with absolute values.
Subsection13.4.1Graphs of Absolute Value Functions
permalinkAbsolute value functions have generally the same shape. They are usually described as “V”-shaped graphs and the tip of the “V” is called the vertex. A few graphs of various absolute value functions are shown in Figure 13.4.2. In general, the domain of an absolute value function (where there is a polynomial inside the absolute value) is (−∞,∞).
Let h(x)=−2|x−3|+5. Using technology, create table of values with x-values from −3 to 3, using an increment of 1. Then sketch a graph of y=h(x). State the domain and range of h.
Let j(x)=||x+1|−2|−1. Using technology, create table of values with x-values from −5 to 5, using an increment of 1 and sketch a graph of y=j(x). State the domain and range of j.
Since the graph of \(y=\abs{x}\) crosses \(y=3\) at the \(x\)-values \(-3\) and \(3\text{,}\) the solution set to the equation \(\abs{x}=3\) must be \(\{-3,3\}\text{.}\)
Since the graph of \(y=\abs{2x+3}\) crosses \(y=5\) at the \(x\)-values \(-4\) and \(1\text{,}\) the solution set to the equation \(\abs{2x+3}=5\) must be \(\{-4,1\}\text{.}\)
Please note that there is a big difference between the expression |3| and the equation |x|=3.
The expression |3| is describing the distance from 0 to the number 3. The distance is just 3. So |3|=3.
The equation |x|=3 is asking you to find the numbers that are a distance of 3 from 0. These two numbers are 3 and −3.
permalinkLet's solve some absolute value equations algebraically. To motivate this, we will think about what an absolute value equation means in terms of the “distance from zero” definition of absolute value. If
|X|=n,
permalinkwhere n≥0, then this means that we want all of the numbers, X, that are a distance n from 0. Since we can only go left or right along the number line, this is describing both X=n as well as X=−n.
Fact 13.4.12 says that the equation \(\abs{x}=6\) is the same as
\begin{equation*}
x=6\text{ or } x=-6\text{.}
\end{equation*}
Thus the solution set is \(\{6,-6\}\text{.}\)
Fact 13.4.12 doesn't actually apply to the equation \(\abs{x}=-4\) because the value on the right side is negative. How often is an absolute value of a number negative? Never! Thus, there are no solutions and the solution set is the empty set, denoted \(\emptyset\text{.}\)
The equation \(\abs{5x-7}=23\) breaks into two pieces, each of which needs to be solved independently.
Let's graphically solve an equation with an absolute value expression on each side: |x|=|2x+6|. Since |x|=3 had two solutions as we saw in Example 13.4.9, you might be wondering how many solutions |x|=|2x+6| will have. Make a graph to find out what the solutions of the equation are.
Figure 13.4.15 shows that there are also two points of intersection between the graphs of \(y=\abs{x}\) and \(y=\abs{2x+6}\text{.}\) The solutions to the equation \(\abs{x}=\abs{2x+6}\) are the \(x\)-values where the graphs cross. So, the solution set is \(\{-6,-2\}\text{.}\)
First break up the equation into the left side and the right side and graph each separately, as in \(y=\abs{x+1}\) and \(y=\abs{2x-4}\text{.}\) We can see in the graph that the graphs intersect twice. The \(x\)-values of those intersections are \(1\) and \(5\) so the solution set to the equation \(\abs{x+1}=\abs{2x-4}\) is \(\{1,5\}\text{.}\)
You might wonder why the negative sign “has” to go on the right side of the equation in X=−Y. It doesn't; it can go on either side of the equation. The equations X=−Y and −X=Y are equivalent. Similarly, −X=−Y is equivalent to X=Y. That's why we only need to solve two of the four possible equations.
Note that one of the two pieces gives us an equation with no solutions. Since \(0\ne3\text{,}\) we can safely ignore this piece. Thus the only solution is \(\frac{1}{2}\text{.}\)
We should visualize this equation graphically because our previous assumption was that two absolute value graphs would cross twice. The graph shows why there is only one crossing: the left and right sides of each “V” are parallel.
Note that our second equation is an identity so recall from Section 2.4 that the solution set is “all real numbers.”
So, our two pieces have solutions \(1\) and “all real numbers.” Since \(1\) is a real number and we have an or statement, our overall solution set is \((-\infty,\infty)\text{.}\) The graph confirms our answer since the two “V” graphs are coinciding.
