$f$ is a function so that $f(ab + 1) = af(b) − f(a) + 6$ for all real numbers $a$ and $b$

Question

If a function $f$ exists so that $f(ab + 1) = af(b) − f(a) + 6$ for all real numbers $a$ and $b$. What are all possible functions $f$ that satisfy this equation and also prove that there are no other functional solutions that make the equation true? Any help is appreciated

Answer 1

Put $b =0$ to get $f(1)=af(0)-f(a)+6$. Solve this for $f(a)$.

You get $f(x)=cx+d$ for some $c$ and $d$. Now go back to the original equation to see what you can say about $c$ and $d$.

The answer is $f(x)=2x+2$.

Answer 2

After Kabo Murphy's "put $b=0$"step, put $a = 0$ and find $f(1) = -f(0) + 6$, allowing you to eliminate $f(1)$ from the two equations, giving $f(a)$ in terms of $f(0)$.

Then put $a = b = 1$ to discover the value of $f(2)$. Use this in your relation $f(a) = \text{[constant expression containing $f(0)$]}$ to find a formula for $f(a)$ in terms of $a$ and constants.