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Linear fractional functions
Example 1
The area of a rectangle is \latex{ 1 } area unit \latex{(}e.g. \latex{ cm^{2}) }, one of its sides has a length of \latex{a\,(cm) }. Give the length of the adjacent side.
Solution
The area of the rectangle is the product of the measure of the two adjacent sides, thus the length of the missing side is \latex{\frac{1}{a} \,(cm,} where \latex{a\gt0)}.
In connection with this we can interpret function \latex{x\mapsto\frac{1}{x}}, \latex{x\gt0} on the set of positive real numbers.
With the help of the defining formula let us extend this function to all the real numbers not equal to \latex{ 0 }:
\latex{f:(\R\setminus\lbrace{0\rbrace})\rightarrow \R}, \latex{f(x)=\frac{1}{x}}.
Let us plot the graph of and characterise the resulting function \latex{ f }. First it is worth realising that if \latex{x\neq 0}, then \latex{f(-x)=\frac{1}{-x}=-\frac{1}{x}=-f(x)}.
DEFINITION: The functions, where \latex{f(–x)=-f(x)} is fulfilled for every \latex{ x } of the domain, are called odd functions.
Since the reflection of point \latex{ P(a; b) } about the origin is \latex{ P’(–a; –b) }, the graph of an odd function is symmetric about the origin. (Figure 56)
So it is enough to plot the image of f when \latex{ x\gt 0 }, and then reflect the resulting curve about the origin. For example when calculating the value of the function at the following places \latex{\frac{1}{3};\,\frac{1}{2};\, 1;\, 2;\, 3} we get a curve easy to draw, and it should be reflected about the origin. The name of the resulting curve is hyperbola. (Figure 57)
\latex{P(a;b)}
\latex{P'(-a;-b)}
Figure 56
\latex{ b }
\latex{ a }
\latex{ -a }
\latex{ -b }
\latex{ x }
\latex{ y }
\latex{y=\frac{1}{x}}
Figure 57
\latex{ 1 }
\latex{ 1 }
\latex{ -1 }
\latex{ -1 }
\latex{ x }
\latex{ y }
We are going to go into details about the hyperbola as a point set only later.
The function is decreasing both on interval \latex{]-\infty}; \latex{0[} and on interval \latex{]0}; \latex{+\infty[}, it has neither a largest nor a smallest value. Its range consists of all real numbers except for \latex{ 0 }. We can observe the following property of the function at positive \latex{ x } places: in the case of large \latex{ x } places its value is small, accordingly the curve “almost touches” the \latex{ x- }axis.
At small positive \latex{ x } places the value of the function will be greater and greater, accordingly the curve will “almost touch” the \latex{ y- }axis as we move toward the origin.
At small positive \latex{ x } places the value of the function will be greater and greater, accordingly the curve will “almost touch” the \latex{ y- }axis as we move toward the origin.
We drew the curve part belonging to the positive \latex{ x } places again so that the curve part connecting any two of the curve points is below the chord connecting these two points, i.e. the function is convex at positive \latex{ x } places. (Figure 58)
\latex{y=\frac{1}{x}}
Figure 58
\latex{ a }
\latex{ b }
\latex{ x }
\latex{ y }
\latex{ P }
\latex{ Q }
\latex{y=\frac{1}{x}}
\latex{P(a;\frac{1}{a})}
\latex{Q(b;\frac{1}{b})}
\latex{a}
\latex{b}
\latex{\frac{a+b}{2}}
Figure 59
\latex{ A }
\latex{ F }
\latex{ B }
\latex{ x }
\latex{ y }
\latex{ S }
\latex{ R }
We are going to prove this property for the mid-point of the line segment connecting the two points. Let \latex{0\lt a\lt b}, then in the figure mid-point \latex{ F } of line segment \latex{ AB } is at \latex{\dfrac{a+b}{2}.}The length of line segment \latex{ SF } is \latex{\frac{1}{\dfrac{a+b}{2} }=\frac{2}{a+b} }. (Figure 59)
Since \latex{ RF } is the mid-line of trapezium \latex{ ABQP }, thus its length is the arithmetic mean of the length of line segments \latex{ AP } and \latex{ BQ }: \latex{\frac{\dfrac{1}{a}+\dfrac{1}{b} }{2} }. We have to show that \latex{\frac{\dfrac{1}{a}+\dfrac{1}{b} }{2}\gt \frac{2}{a+b} }, or the following which is equivalent with it:\latex{\frac{2}{\dfrac{1}{a}+\dfrac{1}{b} } \lt \frac{a+b}{2} }.
A simpler form of the left side: \latex{\frac{2ab}{a+b}}. The right side is greater than the left side if and only if their difference is positive: \latex{\frac{a+b}{2}-\frac{2ab}{a+b}=\frac{(a+b)^{2}-4ab}{2(a+b)}=\frac{(a-b)^{2}}{2(a+b)} \gt 0.}
A simpler form of the left side: \latex{\frac{2ab}{a+b}}. The right side is greater than the left side if and only if their difference is positive: \latex{\frac{a+b}{2}-\frac{2ab}{a+b}=\frac{(a+b)^{2}-4ab}{2(a+b)}=\frac{(a-b)^{2}}{2(a+b)} \gt 0.}
The steps can be reversed, thus the original statement is also true.
Example 2
Let us sketch the graphs of and characterise the following functions which map from the subset, restricted by the given restrictions, of the set of real numbers to the set of real numbers:
\latex{f(x)=\frac{1}{x-3} , (x\neq 3)}; \latex{g(x)=\frac{1}{x+2}-1 , (x\neq -2)};
\latex{h(x)=\frac{1}{1-x}, (x\neq 1)}; *\latex{k(x)=\frac{x-2}{x-3}, (x\neq 3)}.
Solution
From the definition of \latex{f} it can be seen that \latex{f} takes the same value at a place \latex{ 3 } greater than function \latex{x\mapsto \frac{1}{x}} does, thus the image of function \latex{x\mapsto \frac{1}{x}} should be translated by \latex{ 3 } units to the positive direction along the \latex{ x- }axis. Function \latex{ f } is decreasing both on the interval \latex{]–\infty};\latex{3[} and on the interval \latex{]3; +\infty[}, its range consists of all real numbers except for \latex{ 0 }. (Figure 60)
We can get the graph of function \latex{g} from the graph of function \latex{x\mapsto \frac{1}{x}}, so that we translate it by \latex{ 2 } units into the negative direction along the \latex{ x- }axis, and then by \latex{ 1 } unit into the negative direction along the \latex{ y- }axis. (Figure 61)
Figure 60
\latex{y=\frac{1}{x}}
\latex{y=\frac{1}{x-3}}
\latex{ 1 }
\latex{ 1 }
\latex{ 3 }
\latex{ x }
\latex{ y }
It is practical to transform the formula giving the definition of function \latex{ h } as follows:
\latex{h(x)=\frac{1}{1-x}=-\frac{1}{x-1}}, \latex{x\neq1}.
From this it can be seen that the values of \latex{h(x)} are \latex{ –1 } times the values of the function \latex{x\mapsto \frac{1}{x-1}}, so its graph should be reflected about the \latex{ x- }axis. The function is increasing both on the interval \latex{]–\infty; 1[} and on the interval \latex{]1; +\infty[} , its range is the set of real numbers except for \latex{ 0 }. (Figure 62)
Figure 61
\latex{y=\frac{1}{x+2}-1}
\latex{y=\frac{1}{x}}
\latex{ -2 }
\latex{ 1 }
\latex{ 1 }
\latex{ -1 }
\latex{ x }
\latex{ y }
\latex{y=\frac{1}{1-x}}
\latex{y=\frac{1}{x}}
Figure 62
\latex{ 1 }
\latex{ y }
\latex{ x }
\latex{y=\frac{1}{x}}
\latex{y=\frac{x-2}{x-3}}
Figure 63
\latex{ 1 }
\latex{ 3 }
\latex{ x }
\latex{ y }
It is worth transforming the formula giving the definition of function \latex{ k } as follows:
\latex{k(x)=\frac{x-2}{x-3}=\frac{x-3+1}{x-3}=1+\frac{1}{x-3}}, \latex{x\neq3}.
According to this the graph of \latex{ k } can be seen in Figure 63.
Example 3
Let us plot the graphs of the following functions:
\latex{f:(\R \setminus \left\{0\right\} )\longrightarrow \R}, \latex{f(x)=\frac{1}{\mid x\mid}};
\latex{g:(\R \setminus \left\{-2;2\right\} )\longrightarrow \R}, \latex{g(x)=\frac{\mid x\mid-1}{\mid x\mid-2}};
\latex{g:(\R \setminus \left\{-2;2\right\} )\longrightarrow \R}, \latex{g(x)=\frac{\mid x\mid-1}{\mid x\mid-2}};
\latex{h:(\R \setminus \left\{-2;2\right\} )\longrightarrow \R}, \latex{h(x)={\begin{vmatrix}\dfrac{\mid x\mid-3}{\mid x\mid-2}\end{vmatrix}}}.
\latex{y=\frac{1}{\mid x\mid}}
\latex{ y }
\latex{ x }
Figure 64
Solution
Function \latex{ f } is even, because \latex{f(–x) = f(x)}. At positive \latex{x} values \latex{f(x)=\frac{1}{x}}, so it is easy to draw the graph of the function. (Figure 64)
The function \latex{ g } is also even, so it is enough to plot it when \latex{x\geq0}, \latex{x\neq 2}, and then to reflect the resulting graph about the \latex{ y- }axis.
If \latex{x\geq0}, \latex{x \neq2}, then
The function \latex{ g } is also even, so it is enough to plot it when \latex{x\geq0}, \latex{x\neq 2}, and then to reflect the resulting graph about the \latex{ y- }axis.
If \latex{x\geq0}, \latex{x \neq2}, then
\latex{ g(x)=\frac{x-1}{x-2}=\frac{x-2+1}{x-2}=1+\frac{1}{x-2}}. (Figure 65)
Now we reflect the resulting figure about the \latex{ y- }axis, this will be the image of the function \latex{ g }. (Figure 66)
Function \latex{ h } is also even, so it is enough to plot the curve part belonging to the values \latex{x\geq0}, \latex{x\neq 2}, and then to reflect it about the \latex{ y- }axis. Then
Function \latex{ h } is also even, so it is enough to plot the curve part belonging to the values \latex{x\geq0}, \latex{x\neq 2}, and then to reflect it about the \latex{ y- }axis. Then
\latex{h(x)={\begin{vmatrix}\dfrac{x-3}{x-2}\end{vmatrix}}=\begin{vmatrix}1-\dfrac{1}{x-2}\end{vmatrix}}, \latex{(x\geq0}, \latex{x\neq2)}.
First we draw the image of the function \latex{x\mapsto1-\frac{1}{x-2}}, \latex{x\geq0}, \latex{x\neq2 }.
Figure 65
\latex{y=1+ \frac{1}{x-2}}
\latex{ y }
\latex{ x }
\latex{ 1 }
\latex{ 2 }
\latex{ 1 }
Figure 66
\latex{y=\frac{\begin{vmatrix}x\end{vmatrix}-1}{\begin{vmatrix}x\end{vmatrix}-2}}
\latex{ -2 }
\latex{ 2 }
\latex{ x }
\latex{ y }
\latex{ 1 }
\latex{ 1 }
We get its absolute value as follows: where the function is negative, we reflect the curve about the \latex{ x- }axis. The image of \latex{ h } at positive \latex{ x } places is shown in Figure 67.
After this we can draw the image of the whole function. (Figure 68)
Figure 67
\latex{ 1 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ x }
\latex{ y }
\latex{y=\begin{vmatrix}\frac{\mid x\mid -3}{\mid x\mid -2}\\\end{vmatrix}}
Figure 68
\latex{ -3 }
\latex{ -2 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 1 }
\latex{ x }
\latex{ y }
If the area of a rectangle is \latex{ 6 } area units \latex{(}e.g. \latex{ cm^{2}) }, and one of its sides has a length of \latex{ x }, then give the length of the other side.
Since the area is the product of the two sides, the other side is \latex{y=\frac{6}{x}}.
The product of amounts \latex{ x } and \latex{ y } is constant, we say that they are inversely proportional. Inverse proportionality has a close connection to function \latex{f:\R^{+}\rightarrow\R}, \latex{f(x)=\frac{a}{x}} \latex{(a} is constant\latex{)}. We can also say that function \latex{ f } is a function describing the inverse proportionality.
isotherm belonging to the absolute temperature \latex{ T }.
\latex{ T_{1} }
\latex{ T_{2} }
\latex{T_{1}\lt T_2}
\latex{ V }
\latex{ p }
⯁ ⯁ ⯁
We can solve the following exercise about speed with the help of our knowledge about functions.
Example 4
A cyclist has ridden on the road between town \latex{ A } and town \latex{ B } there and back. We know that he was riding his bicycle at a speed of \latex{ 20 } \latex{ \frac{km}{h} } from \latex{ A } to \latex{ B }; about his way back we only know that his speed was at least \latex{ 20 } \latex{ \frac{km}{h} } and at most \latex{ 60 } \latex{ \frac{km}{h} } . What are the limits of his average speed throughout the whole cycling there and back?
Solution
Let \latex{d} denote the distance between \latex{ A } and \latex{ B }, and for ease of calculation let \latex{ 10x } represent the speed of the cyclist on his way back. So the value of \latex{ x } can vary between \latex{ 2 } and \latex{ 6 }.
The time needed to ride from \latex{ A } to \latex{ B } is \latex{\frac{d}{20}}, the time needed to ride from \latex{ B } to \latex{ A } is \latex{\frac{d}{10x}}, so the cyclist needed \latex{\frac{d}{20}+\frac{d}{10x}} to ride the whole road \latex{ 2 }\latex{ d }.
So his average speed was:
So his average speed was:
\latex{s=\dfrac{2d}{\dfrac{d}{20}+\dfrac{d}{10x}}} \latex{(2\leq x\leq 6)}.
Figure 69
\latex{ f }
\latex{ -1 }
\latex{ 1 }
\latex{ 2 }
\latex{ 6 }
\latex{ 1 }
\latex{ x }
\latex{ y }
Let us transform the resulting expression. Let us simplify by \latex{ d }, and then let us multiply both the numerator and the denominator by \latex{ 10 }:
\latex{s=10\times \frac{2}{\dfrac{1}{2}+\dfrac{1}{x}}=10\times \frac{4x}{x+2}=40\times \frac{x+2-2}{x+2}=40\times \biggl(1-\frac{2}{x+2}\biggr)}.
It is enough to determine the largest and the smallest value of function
\latex{g:\lbrack2;6\rbrack\rightarrow\R}, \latex{g(x)=1-\frac{2}{x+2}}. Let us extend function \latex{ g } with the formula to the set of real numbers not equal to \latex{ –2 }, let us denote this function by \latex{ f }, and let us sketch the graph of function \latex{ f }. (Figure 69)
\latex{g:\lbrack2;6\rbrack\rightarrow\R}, \latex{g(x)=1-\frac{2}{x+2}}. Let us extend function \latex{ g } with the formula to the set of real numbers not equal to \latex{ –2 }, let us denote this function by \latex{ f }, and let us sketch the graph of function \latex{ f }. (Figure 69)
Function\latex{ f } is increasing on interval \latex{[2; 6]}, thus its smallest value is \latex{f(2)=\frac{1}{2}}, and its largest value is \latex{f(6)=\frac{3}{4}}. The same places and values will be the extreme places and values of function \latex{ g }. So the average speed of the cyclist for the whole ride was:
\latex{ 20\,\frac{km}{h}\leq s \leq30\,\frac{km}{h}}.
Exercises
{{exercise_number}}. Plot the graphs of and characterise the following functions.
- \latex{f: (\R\setminus\left\{0\right\} )\rightarrow \R, f(x)=-\frac{1}{x}}
- \latex{g: (\R\setminus\left\{4\right\} )\rightarrow \R, g(x)=\frac{1}{x-4} }
- \latex{h: (\R\setminus\left\{2\right\} )\rightarrow \R, h(x)=\frac{1}{2-x} }
- \latex{k: (\R\setminus\left\{-3\right\} )\rightarrow \R, k(x)=\frac{1}{x+3}}
{{exercise_number}}. Plot the graphs of and characterise the following functions.
- \latex{f(x)=\frac{2x-3}{x-2} ,\, x\neq 2}
- \latex{g(x)={\begin{vmatrix}\dfrac{x-4}{x-5}\end{vmatrix}} ,\, x\neq 5}
- \latex{h(x)=\frac{x^{2}-6x+5}{x^{2}-2x+1} ,\, x\neq 1}
- \latex{k(x)=\frac{3\mid x\mid-2}{\mid x\mid-1} ,\, \mid x\mid \neq 1}
{{exercise_number}}. Which of the following amounts are inversely proportional and which are not?
- The speed and time needed to cover a given distance.
- The area and the length of the side of a square.
- The amount of goods (for example flour) and the money paid for it.
- The length of two adjacent sides of a rectangle with fixed area.
{{exercise_number}}. Sketch the graphs of functions \latex{f} and \latex{g} in one coordinate system, and solve equation\latex{ f(x) = g(x) } on the given set.
\latex{f:\lbrack 1;+\infty\lbrack\rightarrow\R,} \latex{f(x)=\sqrt{x+2\sqrt{x-1} } +\sqrt{x-2\sqrt{x-1} } }
\latex{g:\lbrack1;+\infty\lbrack\rightarrow \R,} \latex{g(x)=\frac{1}{x-1}}