Black book 3

$f(42)-f(6)=6\\ f(6)-f(\dfrac { 6 }{ 7 } )=\dfrac { 6 }{ 7 } \\ f(\dfrac { 6 }{ 7 } )-f(\dfrac { 6 }{ { 7 }^{ 2 } } )=\dfrac { 6 }{ { 7^{ 2 } } } \\ \vdots \\ \vdots$

adding upto infinity

$\Rightarrow f(42)-f(\dfrac { 6 }{ { 7 }^{ \infty } } )=6+\dfrac { 6 }{ { 7 }^{ 1 } } +\dfrac { 6 }{ { 7 }^{ 2 } } +\dfrac { 6 }{ { 7 }^{ 3 } } +\dfrac { 6 }{ { 7 }^{ 4 } } \dots \infty \\ \Rightarrow f(42)-f(0)=6+1\\ \Rightarrow f(42)=6+1+1=\boxed{\color{#D61F06}{8}}$

For $f(x) - f \left(\dfrac{x}{7} \right) = \dfrac{x}{7}$ , $f(x)$ must be of the first degree and since $f(0) = 1$ , $f(x)$ must be of the form $f(x) = ax + 1$ . Then, we have:

$\begin{aligned} f(x) - f \left(\frac{x}{7} \right) & = \frac{x}{7} \\ \Rightarrow ax+1 - \frac{ax}{7} - 1 & = \frac{x}{7} \\ ax \left(1-\frac{1}{7}\right) & = \frac{x}{7} \\ a\left(\frac{6}{7}\right) & = \frac{1}{7} \\ \Rightarrow a & = \frac{1}{6} \\ \Rightarrow f(x) & = \frac{x}{6} + 1 \\ f(42) & = \frac{42}{6} + 1 = \boxed{8} \end{aligned}$

f(x)= x/6 + 1.