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Linear functions
Example 1
A square with side \latex{ a } (e.g. \latex{ a } \latex{ cm } when measured in \latex{ centimetre }) is given.
Let us give its perimeter.
Figure 9
\latex{ a }
\latex{ a }
Solution
The perimeter can obviously be expressed as follows: \latex{P = 4a} (\latex{ cm }). (Figure 9)
Let us analyse and interpret the result. \latex{a} is the length of the side of the square measured in \latex{ centimetre }, it can be any positive number, thus the perimeter of the square is four times this value. With the help of this formula we can interpret a function on the set of positive real numbers.
Let the name of the function be \latex{ P }, and the fact that it maps from \latex{\R^{+}} to \latex{\R} is denoted as: \latex{\R^{+}\rightarrow\R}. If \latex{a \in \R}, then the value of \latex{ P } at the place of \latex{ a }, i.e. \latex{P(a) = 4a}.
Shortly: \latex{P: \R^{+}\rightarrow \R, P(a) = 4a}.
Instead of \latex{P(a) = 4a} we can write \latex{a\mapsto 4a}.
Let the name of the function be \latex{ P }, and the fact that it maps from \latex{\R^{+}} to \latex{\R} is denoted as: \latex{\R^{+}\rightarrow\R}. If \latex{a \in \R}, then the value of \latex{ P } at the place of \latex{ a }, i.e. \latex{P(a) = 4a}.
Shortly: \latex{P: \R^{+}\rightarrow \R, P(a) = 4a}.
Instead of \latex{P(a) = 4a} we can write \latex{a\mapsto 4a}.
⯁ ⯁ ⯁
The domain of function \latex{ P } is \latex{\R^{+}}, the set of positive real numbers. \latex{\R} is a codomain of \latex{ P }. The set of elements of \latex{\R} which are outputs of \latex{ P }, i.e. the range of \latex{ P } is also set \latex{\R^{+}}.
DEFINITION: If two non-empty sets are given, sets \latex{ A } and \latex{ B }, and every element of set \latex{ A } relates to an element of set \latex{ B }, then this relation is called a function. The domain of the function is \latex{ A }, its codomain is \latex{ B }. The range of the function is the subset of set \latex{ B } the elements of which are related to. We say the function maps from \latex{ A } to \latex{ B }.
Notation: \latex{f : A\rightarrow B.} If \latex{a \in A,} then \latex{f (a) \in B} denotes the value of function \latex{f} at \latex{a}, in other words its substitution value.
Figure 10
\latex{ 1 }
\latex{ 4 }
\latex{ P }
\latex{ y }
\latex{ x }
\latex{ 1 }
The image or graph of the function \latex{ P } is derived by plotting points \latex{(a;\,4a)} in the Cartesian coordinate system, thus we get a ray (Figure 10).
DEFINITION: If function \latex{f: A \rightarrow B} is given, then the graph (image) of the function is the set of points \latex{ P } in the Cartesian coordinate system, the first coordinate of which is \latex{ a }, and the second coordinate of which is \latex{f(a) \,\,(a\in A; f(a)\in B)}.
The graph of function f can concisely be written as:
{\latex{{P(a; f(a)) \mid a \in A; f(a) \in B}}}.
{\latex{{P(a; f(a)) \mid a \in A; f(a) \in B}}}.
DEFINITION: Given a function \latex{f} any \latex{x} for which \latex{f(x)=0} is called a zero of \latex{f}.
Let us consider the function \latex{g: \R\rightarrow\R,\, g(x) = 4x}.
The domain of \latex{ g } is the set of all real numbers. Its codomains and its range are also the set of all real numbers. Its graph is a straight line passing through the origin. (Figure 11)
The graph of the function \latex{ P } is only a part of \latex{ g }, the open ray (without the end-point) that starts from the origin and is in the first quadrant. We say that function \latex{g} is an extension of function \latex{ P }, and the other way round, \latex{ P } is the restriction of function \latex{ g }. The two functions differ because \latex{ g } has a wider domain since \latex{ \R^{+} \subset \R}, but the assignment rule was given with the same formula for both functions.
Figure 11
\latex{y=4x}
\latex{ 1 }
\latex{ 4 }
\latex{ x }
\latex{ y }
\latex{ 1 }
Example 2
One of the sides of a rectangle is \latex{ 1 } \latex{ cm }; its other side is \latex{ a } \latex{ cm } long. Let us give its perimeter.
\latex{ 1 } \latex{ cm }
\latex{ a } \latex{ cm }
Figure 12
Solution
The perimeter of the rectangle: \latex{P = 2a + 2}. (Figure 12)
In connection with this we can interpret function \latex{P: \R^{+}\rightarrow \R, P(a) = 2a + 2}. The graph of function \latex{ P } is an open ray starting at point (\latex{ 0; 2 }). (Figure 13)
The extension of function \latex{ P } to all real numbers is function \latex{ g }:
Figure 13
\latex{(0;2)}
\latex{ 1 }
\latex{ 4 }
\latex{ P }
\latex{ y }
\latex{ x }
\latex{ 1 }
\latex{g: \R\rightarrow\R,\;\;\; g(x) = 2x + 2.}
The graph of function \latex{ g } is the following straight line. (Figure 14)
Function \latex{ g } can be given in a shorter form as: \latex{g(x) = 2x + 2}, and it is defined so that the domain of function \latex{ g } is the set of real numbers for which the formula has a meaning, so in this case all real numbers.
In the following examples and exercises every function is an \latex{\R\rightarrow\R} type function.
Figure 14
\latex{y = 2x + 2}
\latex{ 2 }
\latex{ y }
\latex{ x }
\latex{ 1 }
\latex{ 1 }
Example 3
Plot the graphs of the following functions in the Cartesian coordinate system:
\latex{f(x) = 3x;\;\;\; g(x) = x;\;\;\; h(x) = –2x;\;\;\; k(x) = –x.}
Solution
Based on experience we accept (later on we are going to prove it precisely) that the graph of all these functions is a straight line. A straight line is unambiguously defined by two of its points, so to plot the graph of a function it is enough to calculate the value of the function at two places and to connect the corresponding points with a straight line. It is often practical to substitute \latex{ 0 } and \latex{ 1 } for \latex{x} in each case, but any \latex{ 2 } values will work. We drew the graphs in one figure. (Figure 15)
The common property of the previous examples is that the assignment rule was given in the form of \latex{x\mapsto mx} (\latex{ m } is a given number). The graph of such functions passes through point (\latex{ 0; 0 }), since their value at \latex{ 0 } is \latex{ 0 }, and it passes through point (\latex{ 1 }; \latex{ m }), since the value of the functions at \latex{ 1 } is \latex{ m }. \latex{ m } typifies the straight line which is the image of the given function. If moved \latex{ 1 } to the right on the \latex{ x }-axis (for example from \latex{ 0 } to \latex{ 1 }), then the \latex{y}-coordinate of the point of the straight line changes by \latex{ m } (from \latex{ 0 } to \latex{ m }), i.e. it increases if \latex{ m\gt 0 }, it decreases if \latex{ m\lt 0 }, and it is constant if \latex{ m = 0 }. m is called the gradient of the straight line.
\latex{y=3x}
\latex{y=-2x}
\latex{y=x}
\latex{y=-x}
Figure 15
\latex{ y }
\latex{ x }
\latex{ 1 }
\latex{ 1 }
Example 4
Let us plot the graphs of the following functions:
\latex{f(x) = 2x + 1; \;\;\; g(x) = –x + 2; \;\;\; h(x) = x\, –\, 3.}
Solution
The image (graph) of all three functions is a straight line, thus it is enough to give \latex{ 2 } points of each graph. We can act in the same way as to calculate the functional values at places \latex{x = 0} and \latex{x = 1}, and then we use the corresponding points to plot the graph. (Figure 16)
Figure 16
\latex{y=x-3}
\latex{y=2x+1}
\latex{(1;3)}
\latex{(1;1)}
\latex{(1;-2)}
\latex{y=x+2}
\latex{ 1 }
\latex{ 2 }
\latex{ -3 }
\latex{y }
\latex{x }
The graph of the function defined by assignment rule \latex{x\mapsto mx+b} (m and b are given numbers) is a straight line, its gradient is m and it intersects the y-axis at point \latex{(0; b)}.
DEFINITION: Functions with the form of \latex{f(x) = mx + b} ( \latex{m\neq} \latex{ 0 } and \latex{ b } are given numbers) are called polynomial functions of degree one. If \latex{ m = 0 }, then the function is given in the form of \latex{ f(x) = b } (constant function), its image is a straight line parallel with the \latex{ x }-axis. The common name of such functions is: linear functions.
Function \latex{P: \R^{+} \rightarrow \R,\;P(a) = 4a} from the first example can also be characterised so that the value of \latex{ P } is always four times the value of \latex{ a }, in other words the quotient of \latex{ P } and \latex{ a } is a constant, \latex{ 4 }. This relation of amount \latex{ P } and \latex{ a } is expressed in another way too: the two amounts are in direct proportion or are directly proportional.
In general we can also say the following about all functions \latex{f: H\rightarrow\R, f (x) = mx \;\;(H\subset \R,)}: it is a function describing direct proportion with \latex{m} as the constant of proportionality.
A very important physical example of direct proportion is the displacement- time function of steady linear motion. If the velocity (speed) of the motion is v (constant), then the displacement is: \latex{s(t) = v × t}. There is a similar relation between the velocity of motion with constant acceleration and the time: \latex{v(t) = a × t} (\latex{ a } is a constant).
Example 5
A truck sets off from Bristol to Manchester at \latex{ 6 } o’clock in the morning and it covers \latex{ 60 } \latex{ km } \latex{ per } \latex{ an } \latex{ hour. } A passenger car sets off from Bristol to Manchester too, but only at \latex{ 7 } o’clock and with a speed of \latex{ 80 } \latex{ \frac{km}{h} } . When does the passenger car catch up with the truck?
Solution
Let us plot the distance (displacement) covered by the two vehicles against time in a common coordinate system. On axis \latex{t} we measure the time in \latex{ hour } that has passed since the setting off of the truck, and on axis \latex{s} we measure the covered distance. Let one unit be equal to \latex{ 60 } \latex{ km } this time to simplify calculation. Then the displacement-time function of the truck can be given as: \latex{f (t) = t, \,t \geq 0}. (Figure 17)
passenger
car
car
Figure 17
\latex{s= \frac{4}{3}t- \frac{4}{3}}
\latex{s=t}
truck
\latex{ 4 }
\latex{ 1 }
\latex{ 2 }
\latex{ 3 }
\latex{ 4 }
\latex{ s }
\latex{ t }
\latex{ 1 }
Since the speed of the passenger car is times the speed of the truck, but it sets off \latex{ 1 } \latex{ hour } later, the graph of the function representing its displacement starts at point (\latex{ 1; 0 }) of the \latex{ t }-axis and has a form of \latex{g(t)=\frac{4}{3}t+a}. Since \latex{g(1) = 0}, then \latex{a=-\frac{4}{3}}, which means that the formula defining function \latex{ g } is:
\latex{g(t)=\frac{4}{3}t-\frac{4}{3},\, t\geq 1}.
According to the figure the graphs of the two functions intersect each other at \latex{ t = 4 }. We check the resulting value: \latex{ f(4) = 4 }, \latex{g(4)=\frac{16}{3}-\frac{4}{3}=4,\, t= 4} it is correct.
Therefore the passenger car catches up with the truck in \latex{ 3 } \latex{ hours }.
Therefore the passenger car catches up with the truck in \latex{ 3 } \latex{ hours }.
Exercises
{{exercise_number}}. Plot the graphs of the following functions defined on the set of real numbers.
- \latex{f (x) = -x + 1}
- \latex{g(x)=-\frac{1}{2}x }
- \latex{h(x)=-2x+\frac{3}{2}}
- \latex{k(x)=\frac{2}{3}x-\frac{4}{3}}
{{exercise_number}}. Give the linear function the graph of which passes through the two given points. Give the gradient of the functions and the point where the graph intersects the \latex{ y }-axis.
- \latex{ P } \latex{ (1; 1) } and \latex{ Q } \latex{ (3; 2) }
- \latex{ P } \latex{ (1; –1) } and \latex{ Q } \latex{ (4; –2) }
{{exercise_number}}. Decide whether the given points lie in the given straight line.
- The points are: \latex{ P_{1}(2;3),P_{2}(3;2),P_{3}(4;7) }; the straight line is the image of function \latex{f :\R\rightarrow\R,\, f (x) = 2x - 1}.
- The points are: \latex{ Q_{1}(1;1),Q_{2}(2;2),Q_{3}(3;-4) }; the straight line is the image of function \latex{g:\R\rightarrow\R,\,g (x) = -2x + 2}.
{{exercise_number}}. Two towns, \latex{ A } and \latex{ B } are \latex{ 200 } \latex{ kilometres } away from each other on the bank of a river. Two ships depart at once, one from \latex{ A } to \latex{ B }, the other one from \latex{ B } to \latex{ A }. The first one covers the distance between \latex{ A } and \latex{ B } in \latex{ 5 } \latex{ hours }, the second one in \latex{ 10 } \latex{ hours }. What time do they meet after their departures? Solve the exercise with the help of functions.