Question

If $f\left( xy \right)=yf\left( x \right)+xf\left( y \right)$ for real values of $x$ and$f\left( 2 \right)=9$, then find$f\left( 4 \right)$. $$$$

Answer 1

We divide both sides of the given functional equation in $f$ by $xy$ and define the function $g\left( x \right)=\dfrac{1}{x}$. We see the obtained functional equation in $g$ is Cauchy’s logarithmic functional whose solutions are $g\left( x \right)=C\log x$ with $C$ as an arbitrary constant. We find $f$, put the given value $f\left( 2 \right)=9$ to get $C$ and then put $x=4$ in the definition of $f$ to get $f\left( 4 \right)$.$$$$

Complete step by step answer:
We know that a functional equation is an equation involving only function. We know from Cauchy type equations also called Cauchy functional equations are given with two binary operations ${{o}_{1}},{{o}_{2}}$ (also called linear maps defined on rational number set $f:Q\to Q$) is defined as
$f\left( x{{o}_{1}}y \right)=f\left( x \right){{o}_{2}}f\left( y \right)$
We can extend the definition of Cauchy’s function equation for the extended domain and range of the function $f$ to real number set that is $f:R\to R$ if $f$ is continuous, monotonic on any interval, bounded on some interval and returns positive value for $x > 0$ which means $x\ge 0\Rightarrow f\left( x \right)\ge 0$.
The solutions of Cauchy’ equations are called additive functions .If we choose ${{o}_{1}}$as multiplication and ${{o}_{2}}$ we get Cauchy’s logarithmic equation
$f\left( xy \right)=f\left( x \right)+f\left( y \right)$
We know that the solutions of the above equation are given by $f\left( x \right)=C\log x$ where $C$ is an arbitrary constant. Logarithmic function satisfies the conditions (continuous, monotonic on any interval, bounded on some interval and returns positive value for $x > 0$) to satisfy Cauchy’s equation in $R$. We are given in the question the functional equation with function $f$ as
$f\left( xy \right)=yf\left( x \right)+xf\left( y \right)$
Let us divide all the terms in the above step by $xy$ and have,
$\Rightarrow \dfrac{f\left( xy \right)}{xy}=\dfrac{f\left( x \right)}{x}+\dfrac{f\left( y \right)}{y}$
Let us define a function $g=\dfrac{f\left( x \right)}{x}$ and use it in the above step in the above step to transform the equation.
$\Rightarrow g\left( xy \right)=g\left( x \right)+g\left( y \right)$
We see that the above equation is Cauchy’s logarithmic functional equation in $g$ whose solutions are given by $g\left( x \right)=C\log x$. So we have

& g\left( x \right)=\dfrac{f\left( x \right)}{x} \\\ & \Rightarrow f\left( x \right)=xg\left( x \right) \\\ & \Rightarrow f\left( x \right)=Cx\log x \\\ \end{aligned}$$ We are also given in the question the value $f\left( 2 \right)=9$. We put in the above step to find the value of arbitrary content, $$\begin{aligned} & f\left( 2 \right)=C2\log 2 \\\ & \Rightarrow C=\dfrac{f\left( 2 \right)}{2\log 2}=\dfrac{9}{2\log 2} \\\ \end{aligned}$$ So the definition of the function is, $$f\left( x \right)=\dfrac{9x\log x}{2\log 2}$$ **We are asked to find the value of $f\left( 4 \right)$ in the above question. So we put $x=4$ in the definition of function to have the answer as, $$f\left( x \right)=\dfrac{9\times 4\times \log 4}{2\log 2}=\dfrac{36\log 4}{\log {{2}^{2}}}=\dfrac{36\log 4}{\log 4}=36$$.** **Note:** We can alternatively solve using differential calculus but for that we need the value of${{f}^{'}}\left( 1 \right)$. We note that as the function from positive real numbers set ${{R}^{+}}$ to real numbers set $R$ and similarly $f:{{R}^{+}}\to R$. If we take ${{o}_{1}}$ as addition and ${{o}_{2}}$ as multiplication we are going to get Cauchy’ equation $f\left( x+y \right)=f\left( x \right)f\left( y \right)$ and whose solutions are $f\left( x \right)=C{{a}^{x}}$for some non-zero real number $a$. If we take ${{o}_{1}},{{o}_{2}}$ both as multiplication we get Cauchy’s equation $f\left( xy \right)=f\left( x \right)f\left( y \right)$ whose solutions are $f\left( x \right)=C{{x}^{t}}$ where $t={{\log }_{a}}b,f\left( a \right)=b$