1

I tried to find $f(0)$, by taking $x=1$,

$$f(0)=\frac{c}{a+b},$$

Then i deduced that $f(x)$ is linear,

Taking $x=(u+1), af(u)+bf(-u)=(u+1)c$

Similarly, $af(-u)+bf(u)=f(1-u)c$, adding them

$2c/(a+b)=f(u)+f(-u)$

Now take u+1,

$f(u+1)+f(-u-1)=2c/(a+b)$

Equating this with the first,

$f(u+1)-f(u)=f(-u)-f(-u-1)=2c/(a+b)$

this lead me to the conclusion that f(x) is linear,

h-squared
  • 1,333
  • 6
  • 8
  • 1
    If you have shown that $f(x)$ is linear then you know that $f(x)=Mx+B$ for two constants $M,B$. Now you can just solve for $M,B$. – lulu Feb 24 '20 at 12:12
  • But note: the function $F(x)=x^3$ satisfies $F(x)+F(-x)=0=2F(0)$ so I am not convinced you have shown that your function $f(x)$ is linear. – lulu Feb 24 '20 at 12:16
  • $f(x)=\frac c {a+b}$ is a non-zero solution. – Kavi Rama Murthy Feb 24 '20 at 12:19
  • @lulu Notice the condition $a^2\ne b^2$ – Qurultay Feb 24 '20 at 12:19
  • @Qurultay I didn't say that my $F(x)=x^3$ satisfied the OP's functional equation, I am just pointing out that knowing that $F(x)+F(-x)=2F(0)$ does not imply that $F(x)$ is linear, which appears to be a claim that the OP is making. – lulu Feb 24 '20 at 12:20
  • @user728189 From $af(u)+bf(-u)=(u+1)c$ you cannot conclude that $f$ is linear. – Kavi Rama Murthy Feb 24 '20 at 12:20
  • If $x=u+1$ then the functional equation gets us $af(u)+bf(-u)=c$. I don't understand where the $u+1$ term on the right comes from (in your comment). – lulu Feb 24 '20 at 12:23
  • Taking $x=(u+1), af(u)+bf(-u)=(u+1)c, Also af(-u)+bf(u)=f(1-u)c, adding them , 2c/(a+b)=f(u)+f(-u), Now take u+1, f(u+1)+f(-u-1)=2c/(a+b), equating this with the first, f(u+1)-f(u)=f(-u)-f(-u-1)=2c/(a+b), this lead me to the conclusion that f(x) is linear – h-squared Feb 24 '20 at 12:23
  • Did you write your functional equation correctly? – lulu Feb 24 '20 at 12:24
  • It is $cx$ on the right side, – h-squared Feb 24 '20 at 12:25
  • Post edit: most of the comments are now irrelevant. But it is still true that knowing that a function $F(x)$ satisfies $F(x)+F(-x)=2F(0)$ does not imply that $F(x)$ is linear. – lulu Feb 24 '20 at 12:27
  • Check my previous long comment, – h-squared Feb 24 '20 at 12:27
  • No. Please edit your post to indicate your reasoning. Don't leave detailed computations for the comments. – lulu Feb 24 '20 at 12:28
  • I updated my post – h-squared Feb 24 '20 at 12:31
  • Please edit your post for clarity. You don't mean $af(-u)+bf(u)=f(1-u)c$ for example. And once again: knowing that $F(x)+F(-x)=2F(0)$ does not imply linearity. – lulu Feb 24 '20 at 12:31
  • af(x-1)+b(1-x)=cx is the functional equation, i substituted x=u+1, to deduce that f(x) is linear – h-squared Feb 24 '20 at 12:33

1 Answers1

1

Let $x-1=t.$ $$af(x-1)+bf(1-x)=cx \implies af(t)+bf(-t)=c(t+1)~~~(1).$$ Let $t\rightarrow -t$; we get $$af(-t)+bf(t)=c(-t+1)~~~(2)$$ Let $f(t)=X, f(-t)=Y$, then we have $aX+bY=ct+c, aY+bX=-ct+c$ Solve for $X$ $$X=f(t)=c[\frac{1}{a+b}+\frac{t}{a-b}]\implies f(x-1)=c[\frac{1}{a+b}+\frac{(x-1)}{a-b}]$$ Finally $$f(x)=c[\frac{1}{a+b}+\frac{x}{a-b}]$$

amWhy
  • 209,954
Z Ahmed
  • 43,235