A kosarad üres
The absolute value function
First there is a reminder of the definition of the absolute value of a real number.
DEFINITION: The absolute value of a real number is the number itself if it is non-negative, and it is its opposite if it is negative:
\latex{\left|a\right|= \begin{cases}\,\,\,\,\,a,\,\,\text{if}\; a\geq 0 \\ -a, \; \text{if} \; a\lt0\end{cases}}
For example: \latex{\left|2\right|=2; \;\left|0\right|=0;\; \left|-3\right|=3; \;\left|-5.28\right|=5.28}.
Example 1
Let us plot the graph of and characterise the function \latex{f: \R\rightarrow \R, \; f(x)=\left|x\right|}.
Solution
The range of the function is the set of non-negative real numbers. Its value at 0 is 0, it has a positive value at every other place. It means that the function has an extreme value at 0, more precisely it has a minimum, and its minimum value is \latex{ f(0) = 0}. (Figure 18)
\latex{y=\left|x\right|}
Figure 18
\latex{ x }
\latex{ y }
\latex{ 1 }
\latex{ 1 }
On the set of non-negative real numbers, i.e. on interval \latex{[0; + \infty[} the function is strictly monotonically increasing. Precisely it means the following: if \latex{0\leq x_1\lt x_2}, then \latex{f(x_1)\lt f(x_2)}.
On the set of non-positive real numbers, i.e. on interval \latex{]-\infty; 0]} the function is strictly monotonically decreasing, i.e. it is true for it that if \latex{x_1\lt x_2\leq 0}, then \latex{f(x_1)\gt f(x_2)}.
⯁ ⯁ ⯁
Henceforth we deal with \latex{\R\rightarrow\R} type functions.
Example 2
Let us plot the graphs of the following functions:
\latex{f(x)=\left|x-2\right|};
\latex{\;g(x)=\left|x\right|-3}.
Solution
Based on the definition of the absolute value f can also be written as:
The image of \latex{f} can be derived from the graph of \latex{x\mapsto \left|x\right|}, so that we translate (or shift) it by \latex{ 2 } to the right along the x-axis. The image of function \latex{g} results from the graph of \latex{x\mapsto \left|x\right|} so that we translate it by 3 downwards along the \latex{y}-axis. (Figure 19)
\latex{\begin{align*}f(x)=\begin{cases}\,\,\,\,\,\,\,x-2, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\text{if} \;x\geq 2 \\ -(x-2)=-x+2, \;\text{if} \;x\lt 2\end{cases}\end{align*}}
Similarly function \latex{g}: \latex{g(x)=\begin{cases}\,\,\,\,x-3, \; \text{if} \;x\geq 0 \\ -x-3, \;\text{if} \;x\lt 0\end{cases}}The image of \latex{f} can be derived from the graph of \latex{x\mapsto \left|x\right|}, so that we translate (or shift) it by \latex{ 2 } to the right along the x-axis. The image of function \latex{g} results from the graph of \latex{x\mapsto \left|x\right|} so that we translate it by 3 downwards along the \latex{y}-axis. (Figure 19)
\latex{y=\left|x-2\right|}
\latex{y=\left|x\right|}
\latex{y=\left|x\right|-3}
Figure 19
\latex{ 2 }
\latex{ -3 }
\latex{ x }
\latex{ y }
\latex{ 1 }
\latex{ 1 }
Example 3
Let us plot the graphs of the following functions:
\latex{h(x)=\left|x+1\right|};
\latex{k(x)=\left|x\right|+2}.
Solution
We can write function \latex{ h } and \latex{ k } in other form based on the definition similarly to the previous example:
\latex{h(x)=\begin{cases}\,\,\,\,\,x+1,\; \text{if}\; x\geq -1\\ -x-1, \;\text{if}\; x\lt-1\end{cases}} \latex{\;\;k(x)=\begin{cases}\,\,\,\,x+2,\; \text{if}\; x\geq0 \\ -x+2, \;\text{if}\; x\lt0\end{cases}}
In this case we can get the graph of \latex{h} from the graph of \latex{x\mapsto \left|x\right|}, so that we translate it by \latex{ 1 }unit to the left along the \latex{ x }-axis, and the graph of \latex{ k } by translating it by \latex{ 2 } units upwards along the \latex{ y }-axis. (Figure 20)
\latex{y=\left|x\right|+2}
\latex{y=\left|x\right|}
\latex{y=\left|x+1\right|}
Figure 20
\latex{ 2 }
\latex{ y }
\latex{ x }
\latex{ -1 }
\latex{ 1 }
\latex{ 1 }
Example 4
Let us plot and characterise the graphs of the following functions:
\latex{f(x)=\Big|\left|x\right|-2\Big|};
\latex{g(x)=\Bigg|\Big|\left|x\right|-2\Big|-1\Bigg|}.
Solution
It is worth producing the graph of function \latex{f} using the following steps. First we plot the image of \latex{ \left|x\right|}, from this one the image of \latex{\left|x\right|-2} is derived by translating it by \latex{2} units in the negative direction along the \latex{ y }-axis (Figure 21). Then the image of the function intersects the \latex{ x }-axis at \latex{ –2 } and \latex{ 2 }. The value of function \latex{\left|x\right|-2} for \latex{-2\lt x\lt 2} is negative, for \latex{x\leq-2} and \latex{x\geq2} it is non-negative. Using this information, and the definition of the absolute value, it is easy to plot the image of \latex{ f }. Where \latex{\left|x\right|-2} is non-negative, the image of \latex{ f } is the same, where \latex{\left|x\right|-2} is negative, there \latex{ f } is multiplied by \latex{ –1 }, which means that this part needs to be reflected about the \latex{ x }-axis. (Figure 22) The function \latex{ f } on the interval \latex{]-\infty; -2]} is decreasing, on the interval \latex{[-2; 0]} it is increasing, then on the interval \latex{[0; 2]} it is decreasing again, and on the interval \latex{[2; +\infty[} it is increasing again. So the function has a minimum at \latex{ –2 } and \latex{ 2 }, and its minimum value is \latex{ 0 }. At \latex{ 0 } the function has a so called local maximum.
\latex{y=\left|x\right|-2}
\latex{y=\left|x\right|}
Figure 21
\latex{ -2 }
\latex{ y }
\latex{ x }
\latex{ 1 }
\latex{y=\Big|\left|x\right|-2\Big|}
\latex{y=\left|x\right|-2}
Figure 22
\latex{ -2 }
\latex{ 2 }
\latex{ 2 }
\latex{ y }
\latex{ x }
\latex{ 1 }
\latex{ 1 }
DEFINITION: We say that function \latex{f} at \latex{ x = a } has a local maximum [minimum], if there is a neighbourhood around point \latex{a} (i.e. there is an interval containing \latex{a} where \latex{a} is the mid-point) in which it is true for every \latex{ x } that \latex{f(a)\geq f(x)} [\latex{f(a)\leq f(x)}].
In this case for example interval \latex{[-2; 2]} is the environment around \latex{0} in which \latex{f(0)=2} is the largest functional value. Interval \latex{[-4; 4]} is also good, but a wider interval is not good, since for example \latex{f(5)=3\gt f(0)=2}.
It is practical to derive the graph of function \latex{g} with the following steps: first we plot the graph of \latex{f} i.e. the graph of the function interpreted with the formula \latex{\left|\left|x\right|-2\right|} (it has already been done), then we plot the graph of function \latex{\Big|\left|x\right|-2\Big|-1}, and finally the graph of function \latex{g}. (Figure 23)
It is practical to derive the graph of function \latex{g} with the following steps: first we plot the graph of \latex{f} i.e. the graph of the function interpreted with the formula \latex{\left|\left|x\right|-2\right|} (it has already been done), then we plot the graph of function \latex{\Big|\left|x\right|-2\Big|-1}, and finally the graph of function \latex{g}. (Figure 23)
\latex{g(x)=\Bigg|\Big|\left|x\right|-2\Big|-1\Bigg|}
Figure 23
\latex{ -3 }
\latex{ -2 }
\latex{ -1 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ y }
\latex{ x }
\latex{ 1 }
In Figure 23 it can be easily seen in which intervals the function \latex{g} decreases, and in which intervals it increases. At \latex{-3;\,-1;\,1;\,3} it has local minima, its minimum value is \latex{0} everywhere, while at places \latex{-2;\,0;\,2} it has local maxima, and its maximum value is \latex{1} everywhere.
Example 5
Let us plot the graph of and characterise function \latex{f(x)=\left|x\right|+\left|x-2\right|}.
Solution
Function \latex{ f } is given with a two-term sum. The first term has a minimum at \latex{ x = 0 }, the second one at \latex{ x = 2 }, so by rewriting the definition of \latex{f}:
\latex{f(x)=\begin{cases}-x-(x-2)=-2x+2, \; \text{if}\; x\lt0 \\ \,\,\,\,\,x-(x-2)=2, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\;\text{if}\; 0\leq x\lt2 \\ \,\,\,\,\,x+x-2\,\,\,\,\,=2x-2, \,\,\,\,\;\text{if}\;2\leq x\end{cases}}
The function \latex{f}, the graph of which can be seen in Figure 24, is decreasing on interval \latex{]–\infty; 0],} it has a constant value of \latex{ 2 } on interval \latex{[0; 2]}, and it is increasing on interval \latex{[2; +\infty[}.
We can say it has a minimum in a wider sense on the whole interval of \latex{[0; 2]}, at all the places of this interval, since the function does not take a smaller value than \latex{ 2 } anywhere.
We can say it has a minimum in a wider sense on the whole interval of \latex{[0; 2]}, at all the places of this interval, since the function does not take a smaller value than \latex{ 2 } anywhere.
Figure 24
\latex{y=\left|x\right|+\left|x-2\right|}
\latex{2}
\latex{2}
\latex{y}
\latex{x}
\latex{1}
\latex{1}
Example 6
Let us plot the graphs of and characterise the following functions:
\latex{f(x)=\left|x+3\right|-\left|x-2\right|};
\latex{g(x)=2\left|x\right|-\left|x-1\right|}.
Solution
It is worth rewriting the definition of function \latex{ f } by splitting it into several cases:
The function is constant on \latex{]–\infty; –3[}, its value is \latex{ –5 }, it is increasing on \latex{[–3; 2[}, and it is constant again on \latex{[2; +\infty[}, its value is \latex{ 5 }.
\latex{f(x)=\begin{cases}\,\,\,\,\,\,\,\,\,-5, \;\text{if}\; x\lt -3 \\ 2x+1, \;\text{if}-3\leq x\lt 2\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,5, \;\text{if}\; 2\leq x \end{cases}}
Based on this we can plot the graph of \latex{f}. (Figure 25)
The function is constant on \latex{]–\infty; –3[}, its value is \latex{ –5 }, it is increasing on \latex{[–3; 2[}, and it is constant again on \latex{[2; +\infty[}, its value is \latex{ 5 }.
Figure 25
\latex{y=\left|x+3\right|-\left|x-2\right|}
\latex{ -5 }
\latex{ 2 }
\latex{ -3 }
\latex{ 5 }
\latex{ y }
\latex{ x }
\latex{ 1 }
\latex{ 1 }
It is again practical to rewrite the assignment rule of function \latex{g} by splitting it into several cases:
\latex{g(x)=\begin{cases}-2x+x-1=-x-1, \text{if}\; x\lt0 \\ \,\,\,\,\,2x+x-1=3x-1, \;\text{if}\; 0\leq x\lt1\\ \,\,\,\,\,2x-x+1=x+1, \,\,\,\;\text{if}\; 1\leq x\end{cases}}
Based on this the graph of the function will be the graph shown in Figure 26.
The monotony of function \latex{g} can easily be seen in the figure. Function \latex{g} has a minimum at \latex{0}, its minimum value is \latex{g(0) = –1}.
Figure 26
\latex{y=2\left|x\right|-\left|x-1\right|}
\latex{ -1 }
\latex{ -1 }
\latex{ 1 }
\latex{ 2 }
\latex{ y }
\latex{ x }
\latex{ 1 }
Example 7
Let us plot the graph of and characterise the following function:
\latex{g(x)=\left|x+3\right|+\left|x\right|+\left|x-2\right|}.
Solution
The terms of the sum defining the function \latex{g} take the value of \latex{ 0 } at places \latex{x=-3, x=0, x=2}, and according to this we can rewrite the definition of \latex{ g } as:
\latex{g(x)=\begin{cases}-3x-1,\; \text{if}\; x\lt-3\\\,\,\,-x+5, \;\text{if}\; -3\leq x\lt0\\ \,\,\,\,\,\,\,\,x+5, \;\text{if}\; 0\leq x\lt 2\\\,\, \,\,\,3x+1, \;\text{if}\; 2\leq x\end{cases}}
The graph of the function is shown in Figure 27. It can easily be seen where the function is increasing and decreasing: it is decreasing on the interval \latex{]–\infty; 0[}, it is increasing on \latex{[0; +\infty[}. At \latex{0} function \latex{g} has a minimum, its minimum value is \latex{ g(0) = 5 }.
Figure 27
\latex{y=\left|x+3\right|+\left|x\right|+\left|x-2\right|}
\latex{ 8 }
\latex{ 7 }
\latex{ 5 }
\latex{ 2 }
\latex{ -3 }
\latex{ y}
\latex{ x}
⯁ ⯁ ⯁
Based on our experience the functions, the assignment rule of which can be given with the sum of expressions in the form of \latex{\left|x-a\right|}, are piecewise linear. It means that the graph of such functions consists of straight line segments and rays.
Example 8
To solve the following fun exercise we use the properties of the absolute value function. We wrote the number of match-sticks on five boxes of matches each (Figure 28). The boxes of matches are placed along the circumference of a circle. We can take over match-sticks from and to adjacent match-boxes. Our goal is to have equal number of match-sticks in every match-box, and that the total number of matchsticks taken over is as little as possible.
Figure 28
Solution
There are \latex{ 80 } match-sticks all together in the match-boxes, so the goal is to have \latex{ 16 } match-sticks in each box. After a few trials it is not complicated to find solutions. But we are looking for the solution where the total number of match-sticks taken over is as little as possible. It would also be good to prove that the method, we found and we assume to mean the least number of takes, indeed has this property.
Let us start up with the fact that in the case of the minimum takes some matchsticks need to be taken over from the box containing \latex{ 24 } match-sticks to the box containing \latex{ 9 }, let us denote this number by \latex{ x }.
Let us start up with the fact that in the case of the minimum takes some matchsticks need to be taken over from the box containing \latex{ 24 } match-sticks to the box containing \latex{ 9 }, let us denote this number by \latex{ x }.
\latex{x-8}
\latex{x}
\latex{x-7}
\latex{x-6}
\latex{x-4}
Figure 29
Now in the box containing \latex{9 + x} match-sticks 16 match-sticks should be left, and \latex{9 + x\, –\, 16 = x \,–\, 7} match-sticks should be taken over to the box containing \latex{ 17 } match-sticks, thus there will be \latex{10 + x} match-sticks in it. Again we leave \latex{ 16 } match-sticks in this, and \latex{10 + x\, –\, 16 = x\, –\, 6} match-sticks should be taken over to the box containing \latex{ 18 } match-sticks, thus there will be \latex{12 + x} matchsticks in it. Again we leave \latex{ 16 } match-sticks in this, and \latex{12 + x\, –\, 16 = x\, –\, 4} match-sticks should be taken over to the box containing \latex{ 12 } match-sticks, thus there will be \latex{x + 8} match-sticks in it. Out of this we take over \latex{x + 8\, –\, 16}, i.e. \latex{x\, –\, 8} match-sticks to the box containing \latex{24 \,–\, x} match-sticks. Thus there will be \latex{24\, –\, x + x\, –\, 8 = 16} match-sticks here too, so we reached our goal. (Figure 29)
We are looking for integer \latex{ x } for which it is true that the total number of takes is the smallest. How shall we count the total number of takes?
If for example we found that \latex{x = 5}, then taking over \latex{x-7 = –\,2} match-sticks obviously means that we have to take over \latex{ 2 } match-sticks into the opposite direction. If we count only the number of takes, then the direction does not matter, obviously \latex{\left|x-7\right|} means the number of match-sticks taken over. Thus the total number of takes is given by the value of the following expression:
We are looking for integer \latex{ x } for which it is true that the total number of takes is the smallest. How shall we count the total number of takes?
If for example we found that \latex{x = 5}, then taking over \latex{x-7 = –\,2} match-sticks obviously means that we have to take over \latex{ 2 } match-sticks into the opposite direction. If we count only the number of takes, then the direction does not matter, obviously \latex{\left|x-7\right|} means the number of match-sticks taken over. Thus the total number of takes is given by the value of the following expression:
\latex{\left| x\right|+\left|x-7\right|+\left|x-6\right|+\left|x-4\right|+\left|x-8\right|}.
Because of the meaning of x this expression now has a meaning for those integers only which satisfy the restriction \latex{-9 \leq x \leq24}. The missing part of the exercise can be formulated as to find the smallest value of the function which maps from the set of integers between \latex{ –9 } and \latex{ 24 } to the set of integers and is defined as follows:
\latex{f(x)=\left| x\right|+\left|x-7\right|+\left|x-6\right|+\left|x-4\right|+\left|x-8\right|}.
Figure 30
\latex{ 25 }
\latex{ 15 }
\latex{ 13 }
\latex{ 12 }
\latex{ 11 }
\latex{ 4}
\latex{ 6}
\latex{ 7}
\latex{ 8}
\latex{ y}
\latex{ x}
\latex{ g}
\latex{ 1}
\latex{ 1}
Let us extend function f to all real numbers, i.e. let us interpret the following function \latex{g}:
\latex{g:\R\rightarrow \R, \;\;g(x)=\left| x\right|+\left|x-7\right|+\left|x-6\right|+\left|x-4\right|+\left|x-8\right|}.
It is not difficult to plot the image of the function since it is the sum of simple expressions with absolute values. (Figure 30)
The minimum of function \latex{ g } is at \latex{ x = 6 }, its minimum value here is \latex{g(6) = 11}. \latex{ 6 } is also in the domain of function \latex{ f }, and since \latex{ f } is the restriction of \latex{ g, f } also has a minimum here, \latex{f(6) = 11}. This result means that with taking over a total of \latex{ 11 } match-sticks we reached our goal, as shown in Figure 31.
The minimum of function \latex{ g } is at \latex{ x = 6 }, its minimum value here is \latex{g(6) = 11}. \latex{ 6 } is also in the domain of function \latex{ f }, and since \latex{ f } is the restriction of \latex{ g, f } also has a minimum here, \latex{f(6) = 11}. This result means that with taking over a total of \latex{ 11 } match-sticks we reached our goal, as shown in Figure 31.
Figure 31
\latex{ 6 }
\latex{ 2 }
\latex{ 2 }
\latex{ 0 }
\latex{ 1 }
In the figure it is shown how many match-sticks need to be taken over from one match-box to which adjacent one. Please note that with the help of the method used for solving this – basically fun – exercise even a lot more complicated e.g. transportation problems can be solved.
Let us also consider the case when someone starts from somewhere else, for example when \latex{ x } denotes the number of match-sticks which need to be taken over from the match-box with \latex{ 17 } match-sticks to the box with \latex{ 9 } match-sticks in the case of the solution with minimum takes. In this case with a calculation very similar to the previous one we get that \latex{x \,–\, 7,\, x + 1,\, x\, –\, 3} and \latex{x\, –\, 1} matchsticks need to be taken over from the match-boxes as shown in the figure. (Figure 32)
The total number of takes here is given by the expression
Let us also consider the case when someone starts from somewhere else, for example when \latex{ x } denotes the number of match-sticks which need to be taken over from the match-box with \latex{ 17 } match-sticks to the box with \latex{ 9 } match-sticks in the case of the solution with minimum takes. In this case with a calculation very similar to the previous one we get that \latex{x \,–\, 7,\, x + 1,\, x\, –\, 3} and \latex{x\, –\, 1} matchsticks need to be taken over from the match-boxes as shown in the figure. (Figure 32)
The total number of takes here is given by the expression
\latex{\left| x\right|+\left|x-7\right|+\left|x+1\right|+\left|x-3\right|+\left|x-1\right|}.
According to this we are looking for the minimum of the function
\latex{h(x) = |x+1| + |x| + |x-1| + |x-3| + |x-7|}
extended to all the real numbers. It takes it at place \latex{x = 1} and its value is \latex{h(1) = 11}, like in the previous case; the necessary takes derive exactly as in the previous case.
It can be seen that whichever take is denoted by \latex{ x }, the minimum of the resulting function will always be \latex{ 11 }.
It can be seen that whichever take is denoted by \latex{ x }, the minimum of the resulting function will always be \latex{ 11 }.
\latex{x+1}
\latex{x-3}
\latex{x-1}
\latex{x}
\latex{x-7}
Figure 32
Exercises
{{exercise_number}}. Plot the graphs of and characterise the following functions defined on the set of real numbers.
- \latex{f(x)=\left|x\right|+1}
- \latex{g(x)=\left|x-1\right|+2}
- \latex{h(x)=2-\left|x-1\right|}
- \latex{k(x)=2\left|x\right|}
- \latex{l(x)=\left|2x\right|}
{{exercise_number}}. Plot the graphs of and characterise the following functions defined on the set of real numbers.
- \latex{f(x)=\left|x\right|+x}
- \latex{g(x)=|x| + |x-3|}
- \latex{h(x)=2\left|x\right|+\left|x+2\right|}
- \latex{k(x)=\left|x+3\right|+\left|x-2\right|}
- \latex{l(x)=\Big|\left|x+3\right|-\left|x-2\right|\Big| }
- \latex{ m(x)=\left|x+4\right|+\left|x-1\right|+\left|x-3\right|}
{{exercise_number}}. Solve the tally of the “match-stick taking over” exercise for the case when
- there are five match-boxes along the circumference of a circle and there are \latex{18;\,14;\, 8;\, 15;\, 5} match-sticks in the match-boxes respectively;
- there are six match-boxes along the circumference of a circle and there are \latex{24,\, 9;\, 26;\, 15;\, 16;\, 30} match-sticks in the match-boxes respectively.
{{exercise_number}}. It was an exercise at a Maths competition:
Let the following functions defined on the set of real numbers be given:
\latex{f:x\mapsto\left|x\right|+\left|x-4\right|,\;\;} \latex{g:x\mapsto\left|x+2\right|+\left|x-2\right|+6}.
Let
\latex{h:x\mapsto\begin{cases}f(x), \;\text{if}\; f(x)\geq g(x)\\ g(x), \;\,\text{if}\;g(x)\gt f(x)\end{cases}}
Plot the graph of \latex{h} on the interval \latex{] –5; 10[}.
