assume $f(4 x-3)+f(3-4 x)=4 x$. find the $f(x)$
I did this:
$\begin{aligned} & t=4 x-3 \\&f(t)+f(-t)=t+3 \\&f(-t)+f(t)=-t+3 \\\Rightarrow &f(t)+f(-t)=3 \end{aligned}$
and stucked here.
Asked
Active
Viewed 80 times
3
-
And what if $f(-t)=-f(t)$ or $f(-t)=f(t)$? – Dr. Sonnhard Graubner Aug 09 '19 at 19:02
-
and $f(-t)=-f(t)$ is impossible! – Dr. Sonnhard Graubner Aug 09 '19 at 19:04
-
@ Dr. Sonnhard Graubner do you mean the question itself could be wrong? – nevermind_15 Aug 09 '19 at 19:06
-
No, only $f(-t)=-f(t)$ is impossible. – Dr. Sonnhard Graubner Aug 09 '19 at 19:07
-
$\sin(-t)=-\sin(t)$. – marty cohen Aug 09 '19 at 19:32
1 Answers
7
There can be no such function $f:\Bbb R\to \Bbb R$. As you already derived, $$ f(t)+f(-t)=t+3 $$ must hold for all $t \in \Bbb R$, and therefore (substitute $t$ by $-t$) also $$ f(-t)+f(t)=-t+3 \, . $$ Subtracting the equations gives $2t = 0$, which obviously cannot hold for all $t \in \Bbb R$.
Or shorter: $$ \text{Set }x = 0 \implies f(-3) + f(3) = 0 \\ \text{Set }x = \frac 32 \implies f(3) + f(-3) = 6 $$
Martin R
- 113,040
-
Sorry, not the original poster, but why can't 2t=0 for all real numbers? – FafaDog Aug 09 '19 at 19:24
-
1