Question

Let $f = \{ (1,1,),(2,3),(0, - 1),( - 1, - 3)\} $ be a function$f(x) = ax + b$ from Z to Z defined by for some integers a, b. Determine a, b.

Answer 1

Hint: We have been given some ordered pair that lies on function f and the function \[f\left( x \right)\]with some unknown constant. Now using those ordered pairs we’ll create the number of equations equivalent to the number of unknown constants by substituting the value in \[f\left( x \right)\].
Now solving those equations we’ll get the required value of the unknown constant.

Complete step-by-step answer:
Given data: $f = \{ (1,1),(2,3),(0, - 1),( - 1, - 3)\} $
$f(x) = ax + b$
Since f is a function containing $(1,1,),(2,3),(0, - 1),( - 1, - 3)$
We can say that $f(1) = 1$
And, $f(2) = 3$
Substituting\[\;x = 1\] in equation(i)
$ \Rightarrow f(1) = a(1) + b$
Substituting the value of $f(1)$
$ \Rightarrow 1 = a + b..........(ii)$
Now, substituting \[x = 2\]in equation(i)
$ \Rightarrow f(2) = a(2) + b$
Substituting the value of $f(2)$
$ \Rightarrow 3 = 2a + b$
$ \Rightarrow b = 3 - 2a$
Now, substituting the value of b in equation(ii)
$ \Rightarrow 1 = a + 3 - 2a$
Solving for like terms
$ \Rightarrow a = 3 - 1$
$\therefore a = 2$
Substituting the value of ‘a’ in equation(ii)
$ \Rightarrow 1 = 2 + b$
$ \Rightarrow 1 - 2 = b$
$\therefore b = - 1$
Therefore the function $f(x) = 2x - 1$
And $a = 2$,$b = - 1$

Note: An alternative method for the above solution can be
Since it is given that the $f(x) = ax + b$, which is an equation of line
And since f is a function containing $(1,1,),(2,3),(0, - 1),( - 1, - 3)$, therefore they are the points lying on the line
Now using the two-point form of a line i.e. if we have two points let say $({x_1},{y_1})$and $({x_2},{y_2})$ that lie on a line then the equation of the line is given by $(y - {y_1}) = \dfrac{{({y_1} - {y_2})}}{{({x_1} - {x_2})}}(x - {x_1})$
Therefore the equation of line passes through $(1,1,)$ and, $(0, - 1)$will be
$ \Rightarrow (y - 1) = \dfrac{{(1 - ( - 1))}}{{(1 - 0)}}(x - 1)$, where $y = f(x)$
$ \Rightarrow (y - 1) = (1 + 1)(x - 1)$
$ \Rightarrow (y - 1) = 2(x - 1)$
Simplifying the brackets
$ \Rightarrow y - 1 = 2x - 2$
$\therefore y = 2x - 1$
$\therefore f(x) = 2x - 1$, which is resulting in the same answer as above