Math in Python: Linear functions

The linear function is the function of the form $f(x) = ax+b,$ where $a,b$ are real numbers and $a\neq 0.$ The function domain is the set of real numbers. A graph of a linear function is called a line. The example of linear function $f(x) = 3x -5$ is shown in following figure.

Figure - The graph of a linear function $f(x) = 3x-5$

Linear function $f(x) = ax+b$ have a zero of the function if and only if the value of the linear function is zero i.e. $f(x) = 0$ $$ f(x) = ax + b = 0 \Rightarrow x = -\frac{b}{a}. The linear function in Cartesian coordinate system is represented as a line for which the $a$ represents the slope and $b$ represents the y-axis intercept value. The linear function $f(x) = ax$ goes through the origin of coordinate system. This is odd function ($f(-x) = f(x)$ i.e. the function is symmetric with respect to origin.

Example 1 - Let's draw a graph of functions $f(x) = 3x$ and $g(x) = 3x-3.$
Solution - To graphically represent those two functions use two positive and two negative $x$ values and 0. $$\begin{eqnarray}\mathrm{For}\quad x = 0:\\ \nonumber f(0)&=& 3\cdot 0 = 0, g(0) = 3\cdot 0 -3 =-3\\ \nonumber\mathrm{For}\quad x = -2: \\ \nonumber f(-2) &=& 3\cdot (-2) = -6, g(-2) = 3\cdot(-2)-3 = -9 \\\nonumber \mathrm{For}\quad x=-1: \\\nonumber f(-1) &=& 3\cdot (-1) = -3, g(-1) = 3\cdot (-1) -3= -6 \\ \nonumber \mathrm{For}\quad x = 1: \\ \nonumber f(1) &=& 3\cdot 1 = 3, g(1) = 3\cdot 1 -3 = 0\\\nonumber \mathrm{For}\quad x=2: \\\nonumber f(2) &=& 3\cdot 2 = 6, g(2) =3\cdot 2 -3 = 3\end{eqnarray}$$ Obtained solutions are shown in following table.

$x$	$-2$	$-1$	$0$	$1$	$2$
$f(x)$	$-6$	$-3$	$0$	$3$	$6$
$g(x)$	$-9$	$-6$	$-3$	$0$	$3$

In following figure the graphical representation of functions $f(x)$ and $g(x)$ are shown.

It is interesting to notice that these two functions are parallelal due to the fact that they have the equal slope.

Example 2 - Let's draw a graph of function $f(x) = -\frac{3}{5}x + 2$
Solution - The linear function can be drawn by determining the values of the function $f(x)$ for two random values of x coordinate. In this example two $x$ values are selected 0 and 3 \begin{eqnarray} f(0) &=& -\frac{3}{5}\cdot 0 + 2 \\ f(0) &=& 2\\ f(3) &=& -\frac{3}{5}\cdot 3 + 2 = -\frac{9}{5} + 2 \\ f(3) &=& \frac{-9+10}{5} = \frac{1}{5} \end{eqnarray}

Example 3 - On following figure a linear function is shown. Find the slope value and y-intercept.

Solution

Example 4 - Let's draw a graph of a function $f(x) = 2$ Determine the y-intercept and slope value ?

Solution

Example 5 - Let's draw a graph of functions $f(x) = x-2$ and $g(x) = -x+2$. Determine when both function increases or decreases.

Solution

Example 6 - Determine a set of solutions for the linear inequality $5x -12 > 0$
Solution - The goal of this example is to see when the linear function is greater than zero. The easiest way is to equate the function with zero. \begin{eqnarray} f(x) &=& 5x - 12 = 0 \\ 5x &=& 12/:5 \Rightarrow x &=& \frac{12}{5}\end{eqnarray} So when x is equal to $\frac{12}{5}$ the value of the function is 0 which means that linear function intersects the x-axis at this point. The good idea would be to calculate the function for two more values where one value is greater than 12/5 and the other is lower than 12/5. \begin{eqnarray}f(3) &=& 5\cdot 3 - 12 = 15 - 12 = 3\\f(2) &=& 5\cdot 2 - 12 = 10 - 12 = -2\end{eqnarray} From obtained values it can be concluded that the function f(x) increases from $x \in \Big\langle \frac{12}{5}, +\infty \Big\rangle$ while decreases for $x \in \Big\langle -\infty, \frac{12}{5}\Big\rangle$
Example 7 - The function shown on following figure have $y$ values of 2 and -1 for which values of $x$.
Solution -
Example 8 - Calculate the output values of $f(x) = \frac{3}{2}x - 5$ for the following values of $x:$ -4,-3,-2, 0, 2, 4, 6, 8 ,and 12. Draw the function.
Solution - \begin{eqnarray}x= -4 \Rightarrow f(-4) &=& \frac{3}{2}(-4) - 5\\f(-4) &=& -11\\x = -3 \Rightarrow f(-3) &=& \frac{3}{2}(-3) - 5 \\f(-3) &=& \frac{-9-10}{2} = -\frac{19}{2}\\ x = -2 \Rightarrow f(-2) &=& \frac{3}{2}(-2) - 5 \\ f(2) &=& -7 \\ x = 0 \Rightarrow f(0) &=& -5 \\ x = 2 \Rightarrow f(2) &=& -2 \\ f(4) &=& \frac{3}{2}\cdot 4 - 5 = 1 \\ x= 6 \Rightarrow f(6) &=& 4 \\ x = 8 \Rightarrow f(8) &=& 7 \\ x = 12 \Rightarrow f(12)&=& 13\end{eqnarray}
Example 9 - For the following functions $f_1(x) = 2x + 8, f_2(x) = -2x+2, f_3(x) = -1.5x +1, f_4(x) = -\frac{2}{3}x - \frac{2}{3}$ determine the function values for $x$ values of -2, 0 ,and 2.
Solution The values of the first function \begin{eqnarray}x = -2 \Rightarrow f_1(-2) &=& 2\cdot(-2) + 8 = 4 \\x = 0 \Rightarrow f_1(0) &=& 2\cdot 0 + 8 = 8\\ x = 2 \Rightarrow f_1(2) &=& 2\cdot 2 + 8 = 12 \end{eqnarray} The values of the second function \begin{eqnarray}x = -2 \Rightarrow f_2(-2) &=& -2\cdot (-2) + 2 = 6\\x = 0 \Rightarrow f_2(0) &=& -2\cdot 0 + 2 = 2\\ x = 2 \Rightarrow f_2(2) &=& -2\cdot 2 + 2 = -2 \end{eqnarray} The values of the third function \begin{eqnarray}x = -2 \Rightarrow f_3(-2) &=& -1.5\cdot(-2) + 1 = 4\\x = 0 \Rightarrow f_3(0) &=& -1.5\cdot 0 + 1 = 1\\ x = 2 \Rightarrow f_3(2) &=& -1.5\cdot 2 + 1 = -2 \end{eqnarray} The values of the fourth function \begin{eqnarray}x = -2 \Rightarrow f_4(-2) &=& -\frac{2}{3}\cdot (-2) - \frac{2}{3} = \frac{2}{3}\\x = 0 \Rightarrow f_4(0) &=& -\frac{2}{3}\cdot 0 - \frac{2}{3} = \frac{2}{3}\\ x = 2 \Rightarrow f_4(2) &=& - \frac{2}{3}\cdot 2 - \frac{2}{3} = -2 \end{eqnarray}
Example 10 - For the following functions $f_1(x) = 3x-2$, $f_2(x) = x-2$, $f_3(x) = x-1$, $f_4(x) = x$ determine the function values for $x$ values of -3, 0, and 3.
Solution - The values of the first function \begin{eqnarray} x = -3 \Rightarrow f_1(-3) &=& 3\cdot(-3) - 2 = -11 \\ x= 0 \Rightarrow f_1(0) &=& 3\cdot 0 - 2 = -2 \\ x = 3 \Rightarrow f_1(3) &=& 3 \cdot 3 -2 = 7\end{eqnarray} The values of the second function \begin{eqnarray} x = -3 \Rightarrow f_2(-3) &=& -3 -2 = -5 \\ x= 0 \Rightarrow f_2(0) &=& 0 - 2 = -2 \\ x = 3 \Rightarrow f_2(3) &=& 3 -2 = 1\end{eqnarray} The values of the third function \begin{eqnarray} x = -3 \Rightarrow f_3(-3) &=& -3 -1 = -4 \\ x= 0 \Rightarrow f_3(0) &=& 0 -1 = -1 \\ x = 3 \Rightarrow f_3(3) &=& 3 - 1 = 2\end{eqnarray} The values of the fourth function \begin{eqnarray} x = -3 \Rightarrow f_4(-3) &=& -3\\ x= 0 \Rightarrow f_4(0) &=& 0 \\ x = 3 \Rightarrow f_4(3) &=& 3\end{eqnarray}
Example 11 - Determine from shown lines in following figure which one represents the $f(x) = 2x-1$ function?

Solution - To determine which one of the shown lines in previous figure represents linear function $f(x) = 2x-1$ the two linear function values should be determined for two values of x. The first one will be the y-intercept value i.e the value of the function when argument (x-value) is 0. \begin{eqnarray}x = 0 \Rightarrow f(0) &=& 2\cdot 0 - 1 = -1 \end{eqnarray} From obtained solutions and the the lines shown in previous figure it can be concluded that the green line is potentially linear function due to the fact that this is the only line in figure that for x value of 0 has the f(0) = -1. For the second linear function value the arugment x is equal to 1. \begin{eqnarray}x = 1 \Rightarrow f(1) &=& 2 \cdot 1 - 1 = 1\end{eqnarray} Based on the obtained solution and the lines shown in previous figure it can definately be concluded that the green line represents the linear function $f(x) = 2x-1 .$

Example 12 - Draw the linear function $f(x) = x - 2$ in Cartesian coordinate system.
Solution - Before drawing the function the values of the function will be calculated for $x$ vallues in range from -5 up to 5 and presented in table form. \begin{eqnarray} x = -5 \Rightarrow f(-5) &=& -5 -2 = -7 \\ x = -4 \Rightarrow f(-4) &=& -4 -2 = -6 \\x = -3 \Rightarrow f(-3) &=& -3 -2 = -5 \\ x = -2 \Rightarrow f(-2) &=& -2 -2 = -4 \\ x = -1 \Rightarrow f(-1) &=& -1 - 2 = -3 \\x = 0 \Rightarrow f(0) &=& -2 \\ x= 1 \Rightarrow f(1) &=& 1-2 = -1 \\ x = 2 \Rightarrow f(2) &=& 2- 2 = 0 \\x =3 \Rightarrow f(3) &=& 3 -2 = 1 \\ x= 4 \Rightarrow f(4) &=& 4-2 = 2 \\ x = 5 \Rightarrow f(5) &=& 5-2 = 3\end{eqnarray} All obtained values are givne in table form.

$x$	-5	-4	-3	-2	-1	0	1	2	3	4	5
$f(x)$	-7	-6	-5	-4	-3	-2	-1	0	1	2	3

Here the output of function $f(x)$ was calculated for multiple values of $x$ and this is not necessary due to the fact that only two points are needed for graphically represented linear funciton. Based on the obtained values the graphical representatioan of the function is shown in following figure.

Example 13 - Determine from the shown lines in following figure which one represents the $f(x) = \frac{1}{2}x -2$

Solution - Let's start by discovering the y-intercept. The line that corresponds to the linear function $f(x) = \frac{1}{2}x -2$ must have y-intercept value of -2 for value of x equal to 0. \begin{eqnarray} x = 0 \Rightarrow f(0) &=& \frac{1}{2}\cdot 0 - 2 = -2 \end{eqnarray} Based on the obtained value of y-itercept and shown lines in the previous figure it can be concluded that only the orange line for x value of 0 has the y value (y-itercept) of -2. Just to be sure the value of liner function will be calcualted for $x$ value of 4. \begin{eqnarray} x = 4 \Rightarrow f(4) &=& \frac{1}{2}\cdot 4 - 2 = 2 -2 = 0.\end{eqnarray} From obtained value and the lines shown in previous figure it can be concluded that the orange line represents the linear function $f(x) = \frac{1}{2} x - 2 $
Example 14 - For functions $f_1(x) = -2x -1 $, $f_2(x) = (2-\sqrt{5})x+\sqrt{5}$, $f_3(x)=4x,$ and $f_4(x) = \frac{2}{7}x-3$ determine the function values for $x$ values of -5, -2, 0, 2, and 4.
Solution - The values of function $f_1(x) = -2x -1 $ for different values of $x$. \begin{eqnarray}x = -5 \Rightarrow f_1(-5) &=& -2\cdot (-5) - 1 = 9 \\x = -2 \Rightarrow f_1(-2) &=& -2 \cdot (-2) - 1 = 3 \\ x= 0 \Rightarrow f_1(0) &=& -2 \cdot 0 - 1 = -1 \\ x = 2 \Rightarrow f_1(2) &=& -2 \cdot 2 -1 = -5 \\x = 4 \Rightarrow f_1(4) &=& -2 \cdot 4 -1 = -9\end{eqnarray} The values of the linear function $f_2(x) = (2-\sqrt{5})x + \sqrt{5}$ for different values of $x$. \begin{eqnarray}x = -5 \Rightarrow f_2(-5) &=& (2-\sqrt{5})\cdot (-5) + \sqrt{5} = \sqrt{5} - 5(2-\sqrt{5}) \approx 3.41641 \\x = -2 \Rightarrow f_2(-2) &=& (2-\sqrt{5})\cdot (-2) + \sqrt{5} = \sqrt{5} - 2(2-\sqrt{5}) \approx 2.7082 \\ x= 0 \Rightarrow f_2(0) &=& (2-\sqrt{5})\cdot 0+ \sqrt{5} = \sqrt {5} \approx 2.23607 \\ x = 2 \Rightarrow f_2(2) &=& (2-\sqrt{5})\cdot 2+ \sqrt{5} \approx 1.76393 \\x = 4 \Rightarrow f_2(4) &=& (2-\sqrt{5})\cdot 4+ \sqrt{5}\approx 1.2918\end{eqnarray} The values of the linear function $f_3(x) = 4x$ for different values of $x$. \begin{eqnarray}\end{eqnarray}
Example 15 Test if linear function $f(x) = - \frac{1}{3}x - 3$ is positive for $x>-10$
Solution - To test if the function is positive of the values of $x$ greater than -10 simply pick a number greater than that value let's say -6 and -3 and calculate the function value. \begin{eqnarray}x = - 6\Rightarrow f(-6) &=& -\frac{1}{3}\cdot (-6) - 3 = -1 \\x = - 3 \Rightarrow f(-3) &=& -\frac{1}{3}\cdot (-3) - 3 = -2\end{eqnarray} Based on the obtained solutions it can be concluded that the function is negative for values greater than -10.
Example 16 Test if linear function $f(x) = 2x -4$ is positive for values of $x> 2$
Solution - To test this hypothesis we will select two values of $x$; 4 and 8 and calculate the $f(x)$ value. \begin{eqnarray}x = 4 \Rightarrow f(4) &=& 2\cdot 4 - 4 = 4 \\x = 8 \Rightarrow f(8) &=& 2\cdot 8 - 4 = 12 \end{eqnarray} From obtained solutions it can be concluded that the function is positive for values x > 2.
Example 17 Test if the linear function $f(x) = -\frac{2}{3}x + \frac{4}{3}$ is negative for $x>2$
Solution - To test this hypothesis two values of $x$ are chosen 3 and 6 and the function value is calculated. \begin{eqnarray}x = 3 \Rightarrow f(3) &=& -\frac{2}{3}\cdot 3 -+ \frac{4}{3} = -2 + \frac{4}{3}\\f(3) &=& \frac{-6 + 4}{3} = -\frac{2}{3} \\ x = 6 \Rightarrow f(6) &=& -\frac{2}{3}\cdot 6 - \frac{4}{3} = \frac{-12 - 4}{3} = -\frac{16}{3} \end{eqnarray} From obtained solutions it can be concluded that the linear function is negative for $x$ values greater than 2.
Example 18 Test if linear function $f(x) = 3x-8$ is positive for x > 2
Solution - To test if the linear function is positive for $x$ values greater than 2 the two $x$ values are selected (3,6) for which the function values will be calculated.\begin{eqnarray}x =3 \Rightarrow f(3) &=& 3\cdot 3 - 8 = 9 -8 = 1 \\ x = 6 \Rightarrow f(6) &=& 3\cdot 6 - 8 = 10 \end{eqnarray} From obtained solutions it can be concluded that the function is positive for values of $x> 2$
Example 19 Four points $A_1(-1,7), A_2(0,4), A_3(1,1), A_4(-2,1)$ and the linear function $f(x) = -3x+4$ are given. Determine which of these points do not lie on line of linear function.
Solution - \begin{eqnarray}A_1(-1,7) \Rightarrow f(-1) &=& 7 \\ -3\cdot(-1) + 4 &=& 7\\ 3 + 4 &=& 7 \Rightarrow 7 = 7 \end{eqnarray} The point $A_1$ is the point of the linear function. Check if the point $A_2(0,4)$ belongs to the line. \begin{eqnarray} A_2(0,4) \Rightarrow f(0) &=& 4 \\ -3\cdot 0 + 4 &=& 4 \Rightarrow 4 = 4\end{eqnarray} The point $A_2(0,4)$ belongs to the line. Check if the point $A_3(1,1)$ belongs to the line. \begin{eqnarray} A_3(1,1) \Rightarrow f(1) &=& 1 \\ -3\cdot 1 + 4 &=& 1 \Rightarrow 1 = 1\end{eqnarray} The point $A_3(1,1)$ also belongs to the line. Check if the point $A_4(-2,1)$ belongs to the line. \begin{eqnarray}A_4(-2,1) \Rightarrow f(-2) &=& 1 \\ -3\cdot (-2) + 4 &=& 1 \Rightarrow 10 \neq 1 \end{eqnarray} The point $A_4(-2,1)$ do not belong to the line.
Example 20 Determine the slope coefficient from formula f(x) = ax + 3
Solution - \begin{eqnarray}\end{eqnarray}
Example 21 Determine the output value of the linear function $f(x) = 2x-5$ if $x=-1.$
Solution - \begin{eqnarray}x=-1 \Rightarrow f(-1) &=& 2\cdot (-1) - 5 = -7\end{eqnarray}
Example 22 The slope $a$ of the line parallel to x-axis is equal to?
Solution The line that is parallel to the x-axis has the slope equal to 0 ($a=0$).
Example 23 The linear function has a slope coefficient of 5 and goes through the point (1,10). Determine the linear function.
Solution - The function can be written as $f(x) = 5x + b$. Since the line goes through the point (1,10) the coefficient $b$ or y-intercept can be found.\begin{eqnarray} 5\cdot (1) + b &=& 10 \Rightarrow b = 5\end{eqnarray} The function can be written as $f(x) = 5x + 5$
Example 24 The cost of renting a car is $ 100 for a deposit and $ 30 per day. What is the function of renting a car?
Solution The $ 100 will represent y-intercept since this is starting vlaue of car rental (fixed value). The $ 30 per day will be written as $$30\cdot d$ where $30 will represent slope coefficient and d will represent number of days. The function of renting a car can be written as \begin{eqnarray} P(d) = 30\cdot d + 100\end{eqnarray}

\(x\)	\(-2\)	\(-1\)	\(0\)	\(1\)	\(2\)
\(f(x)\)	\(-6\)	\(-3\)	\(0\)	\(3\)	\(6\)
\(g(x)\)	\(-9\)	\(-6\)	\(-3\)	\(0\)	\(3\)

Math in Python

Linear functions

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