1
$\begingroup$

I stumbled upon this problem on composite functions in a school magazine under 'Quiz'. The problem says:

You are given the composite function $f(x+f(y))=bx+cy$ where $b$ and $c$ are real numbers, $b\neq-1$ with $x$ and $y$ being real variables.

Find the value of $f(\frac 1b)$ and the relationship between $b$ and $c$.

So far, I have worked out that:
$f(f(y)) = bx+cy-x$
$\therefore f(f(0))=bx-x$ ; calling this equation 1
and
$f(x)=bx+cy-f(y)$
$\therefore f(0)=cy-f(y)$ ; calling this equation 2

Substituting equation 2 into equation 1,
I get: $f(cy-f(y))=bx-x$
but this gets me to an equation similar to the original function.

Have my operations been correct? How do I continue from where I have left off to arrive at the desired answer?

$\endgroup$
  • 1
    $\begingroup$ Your equations are wrong. It seems that you are making confusion between $f(x+f(y))$ and $x+f(f(y))$. $\endgroup$ – Crostul Sep 7 '15 at 12:16
  • 1
    $\begingroup$ I don't understand yuor claim that $f(f(y))=bx+cy-x$. I'd rather say that $f(f(y))=cy$, as this is obtained letting $x=0$ in the given equation. $\endgroup$ – Aretino Sep 7 '15 at 12:16
  • $\begingroup$ @Crostul I see. Thank you. I had assumed that if a function $f(a+b)=c$, then $f(b)=c-a$. My assumption must have been false. $\endgroup$ – ChrisJWelly Sep 7 '15 at 12:24
  • $\begingroup$ @Clayton I do not understand. In the first line of my working, didn't I take $y=0$ on the left-hand side? $\endgroup$ – ChrisJWelly Sep 7 '15 at 12:25
  • $\begingroup$ @ChrisJWelly: I see; I misread your mistake. Crostul is right that you had a misunderstanding with $f(x+f(y))$ and $x+f(f(y))$. $\endgroup$ – Clayton Sep 7 '15 at 12:28
6
$\begingroup$

First, when $y=0$, $f(x+f(0))=bx$

Now setting $x=-f(0)$, we have $f(0)=-bf(0)\Rightarrow f(0)=0$ since $b\neq-1$

Hence $f(x+0)=f(x)=bx\Rightarrow f(\frac 1b)=1$

Now $f(x+f(y))=b(x+f(y))=bx+cy\Rightarrow bf(y)=cy\Rightarrow f(y)=\frac cby=bx$

Therefore, with $y=x$, we have $c=b^2$

$\endgroup$
  • $\begingroup$ In your third line of working, you said that $f(x)=bx\Rightarrow f(\frac 1b)=1$. Why is that so? Is this done through cross-multiplication? In your fourth line of working, why is $f(x+f(y))=b(x+f(y))$? and how was the deduction made that $y=x$? Thank you for your help so far! $\endgroup$ – ChrisJWelly Sep 7 '15 at 12:46
  • $\begingroup$ First question:I have replaced $x$ with $\frac 1b$. $\endgroup$ – David Quinn Sep 7 '15 at 12:48
  • $\begingroup$ second question: it is established that $f(\text{whatever})=b\times\text{whatever}$ $\endgroup$ – David Quinn Sep 7 '15 at 12:50
  • $\begingroup$ third question: I am setting $y=x$ to get this $\endgroup$ – David Quinn Sep 7 '15 at 12:50
  • $\begingroup$ I do not understand the response to the third question. If you are setting $y=x$, does that not mean that it is an assumption that has been made, considering that it was not given in the question and it has not been established in the workings? $\endgroup$ – ChrisJWelly Sep 7 '15 at 12:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.