Suppose $f$ is a real function satisfying $f(x+f(x))$ = $4f(x)$ and $f(1) = 4$. Then the value of $f(21)$?

Question

Should I proceed with just putting the value of $f(1)=4$ in the first equation or there will be a different way of solving this ?

Answer 1

You are right. You aren't given much, so use what you know. Let $x=1$ in the first equation and what do you learn? How does it help?

Answer 2

Substituting $x = 1$ results that $f(5) = 16$. Now substitute $x = 5$.

Answer 3

$$ f(1)=4 \\ f(1+f(1)) = f(5) = 4f(1) = 16, $$ and then $5+f(5)=21$ and so on.

Answer 4

Assuming $f$ invertible, let us set $g=f^{-1}$, and $$f(x+f(x))=4f(x)$$ can be rewritten as $$f(g(y)+y)=4y,$$ then $$g(y)+y=g(4y).$$ Setting $h(k)=g(4^k)$, $$h(k+1)=h(k)+4^k,$$ which, with the initial condition $g(4)=h(1)=1$ is easily solved as $$h(k)=\frac{4^k-1}3,$$ i.e. $$g(y)=\frac{y-1}3,$$ or $$\color{green}{f(x)=3x+1}.$$

As we can check, $$3(x+(3x+1))+1=12x+4=4(3x+1).$$

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More generally, the function can be defined for all reals by setting values arbitrarily in $(0,1]$ and using the fact that $$f(3^kx+\frac{4^k-1}3)=3^kf(x)+\frac{4^k-1}3$$ in other intervals, forming a self-similar plot.

Answer 5

Replace $x$ by $x+f(x)$ in the original equation, which then becomes

$f(x+f(x)+f(x+f(x))) = 4f(x+f(x))$
i.e $f(x+f(x)+4f(x))=4*4f(x)$
i.e $f(x+5f(x))=16f(x)$

Setting $x=1$ and using the value $f(1)=4$ we get $f(1+20)=f(21)=16*4=64$