Question 4: Given that \(f(x)=ax+b+1\), where \(a\) and \(b\) are constant numbers. If \(f(3)=8\) and \(f(6)=14\), then find the values of \(a\) and \(b\).
Solution:
\(f(x)=ax+b+1\)
\(f(3)=3a+b+1\)
\(3a+b+1=8\)\(\ \_\_\_\_\_\_\ (1)\)\(\quad \because f(3)=8\)
\(f(6)=6a+b+1\)
\(6a+b+1=14\)\(\ \_\_\_\_\_\_\ (2)\)\(\quad \because f(6)=14\)
Subtracting equation \((1)\) from \((2)\), we have
\(\ \ 6a+b+1=14\)
\(\underline{\underset{-}{}3a\underset{-}{+}b\underset{-}{+}1=\underset{-}{}8}\)
\(\ \ 3a\ \ \ \ \ \ \ \ \ \ \ \ \ \ =6\)
\(a=\large{\frac{6}{3}}\)
\(a=2\)
Put value of \(a\) in equation \((1)\), we have
\(3(2)+b+1=8\)
\(6+b+1=8\)
\(b+7=8\)
\(b=8-7\)
\(b=1\)