Question
Question: Let \(f = \\{ (1,1,),(2,3),(0, - 1),( - 1, - 3)\\} \) be a function\(f(x) = ax + b\) from Z to Z def...
Let f=(1,1,),(2,3),(0,−1),(−1,−3) be a functionf(x)=ax+b from Z to Z defined by for some integers a, b. Determine a, b.
Solution
We have been given some ordered pair that lies on function f and the function f(x)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(x).
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
Substitutingx=1 in equation(i)
⇒f(1)=a(1)+b
Substituting the value of f(1)
⇒1=a+b..........(ii)
Now, substituting x=2in equation(i)
⇒f(2)=a(2)+b
Substituting the value of f(2)
⇒3=2a+b
⇒b=3−2a
Now, substituting the value of b in equation(ii)
⇒1=a+3−2a
Solving for like terms
⇒a=3−1
∴a=2
Substituting the value of ‘a’ in equation(ii)
⇒1=2+b
⇒1−2=b
∴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 (x1,y1)and (x2,y2) that lie on a line then the equation of the line is given by (y−y1)=(x1−x2)(y1−y2)(x−x1)
Therefore the equation of line passes through (1,1,) and, (0,−1)will be
⇒(y−1)=(1−0)(1−(−1))(x−1), where y=f(x)
⇒(y−1)=(1+1)(x−1)
⇒(y−1)=2(x−1)
Simplifying the brackets
⇒y−1=2x−2
∴y=2x−1
∴f(x)=2x−1, which is resulting in the same answer as above