We want the distance between [latex]x[/latex] and 5 to be less than or equal to 4. We can draw a number line to represent the condition to be satisfied.

The distance from [latex]x[/latex] to 5 can be represented using [latex]|x - 5|[/latex]. We want the values of [latex]x[/latex] that satisfy the condition [latex]|x - 5|\le 4[/latex].

Analysis of the Solution

Note that

[latex]\displaystyle{-4}\le{x - 5}[/latex]
[latex]\displaystyle{1}\le{x}[/latex]
And:
[latex]\displaystyle{x-5}\le{4}[/latex]
[latex]\displaystyle{x}\le{9}[/latex]

So [latex]|x - 5|\le 4[/latex] is equal to [latex]1\le x\le 9[/latex].

However, mathematicians generally prefer absolute value notation.

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 [latex]R[/latex] in ohms,

[latex]|R - 680|\le 34[/latex]

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.

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.

From this information we can write the equation

[latex]\begin{array}{l}f\left(x\right)=2\left|x - 3\right|-2,\hfill & \text{treating the stretch as a vertical stretch, or}\hfill \\ f\left(x\right)=\left|2\left(x - 3\right)\right|-2,\hfill & \text{treating the stretch as a horizontal compression}.\hfill \end{array}[/latex]

Analysis of the Solution

Note that these equations are algebraically the same—the stretch for an absolute value function can be written interchangeably as a vertical or horizontal stretch or compression.

[latex]\begin{array}{l}0=|4x+1|-7\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \text{Substitute 0 for }f\left(x\right).\hfill \\ 7=|4x+1|\hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \text{Isolate the absolute value on one side of the equation}.\hfill \\ \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ 7=4x+1\hfill & \text{or}\hfill & \hfill & \hfill & \hfill & -7=4x+1\hfill & \text{Break into two separate equations and solve}.\hfill \\ 6=4x\hfill & \hfill & \hfill & \hfill & \hfill & -8=4x\hfill & \hfill \\ \hfill & \hfill & \hfill & \hfill & \hfill & \hfill & \hfill \\ x=\frac{6}{4}=1.5\hfill & \hfill & \hfill & \hfill & \hfill & \text{ }x=\frac{-8}{4}=-2\hfill & \hfill \end{array}[/latex]

The function outputs 0 when [latex]x=1.5[/latex] or [latex]x=-2[/latex].