Question

Let $f:R\to R$ be a differentiable function satisfying $f(x+y)=f\left(x\right)+f\left(y\right)+xy$ for all $x,y\in R$ and $\underset{h\to 0}{lim}\frac{1}{h}f\left(h\right)=3.$ If the minimum value of $f\left(x\right)$ is $k$, then the value of $|2k|$ is

Answer 1

$f(x+y)=f\left(x\right)+f\left(y\right)+xy$ for all $x,y\in R\cdots \left(1\right)$
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(x\right)+f\left(h\right)+xh-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(h\right)+xh}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(h\right)}{h}+\underset{h\to 0}{lim}x$
$=3+x$
$\therefore f\left(x\right)=3x+\frac{x^2}{2}+C$

Putting $x=y=0$ in $\left(1\right)$, we get
$f\left(0\right)=2f\left(0\right)\Rightarrow f\left(0\right)=0$
So, $C=0$
Hence, $f\left(x\right)=3x+\frac{x^2}{2}$

Putting $f^{\prime }\left(x\right)=0$, we get
$x=-3$
$f^{\prime \prime }\left(x\right)=1>0$
$\therefore f_{min}=f(-3)=-\frac{9}{2}\Rightarrow |2k|=9$