Question

Let $f\left(\frac{x+y}{2}\right)=\frac{f\left(x\right)+f\left(y\right)}{2}$ and $f^{\prime }\left(0\right)=3$ and $f\left(0\right)=5$. Find $f^{\prime \prime }\left(x\right)$. (where $y$ is independent of $x$), when $f\left(x\right)$ is differentiable.

Answer 1

Given $f\left(\frac{x+y}{2}\right)=\frac{f\left(x\right)+f\left(y\right)}{2}$, for any real $x,y$
Also, given $y$ is independent of $x$. ie, $\frac{dy}{dx}=0$
On differentiating w.r.t. $x$, we get
$f^{\prime }\left(\frac{x+y}{2}\right)\cdot \frac{1}{2}(1+\frac{dy}{dx})=\frac{1}{2}\left\{f^{\prime }\right(x)+f^{\prime }(y)\cdot \frac{dy}{dx}\}$
$\therefore \frac{1}{2}{f}^{\prime }\left(\frac{x+y}{2}\right)=\frac{1}{2}{f}^{\prime }\left(x\right)$ (as $\frac{dy}{dx}=0)$
$\Rightarrow f^{\prime }\left(\frac{x+y}{2}\right)=f^{\prime }\left(x\right)$ ....(i)
Substitute $x=0$ and $y=x$ in equation (i), we get
$f^{\prime }\left(\frac{0+x}{2}\right)=f^{\prime }\left(0\right)$
$\Rightarrow f^{\prime }\left(\frac{x}{2}\right)=a$ (given)
which shows $f\left(\frac{x}{2}\right)=a\left(\frac{x}{2}\right)+c$
$\therefore f\left(x\right)=ax+c$
Substitute $x=0$
$f\left(0\right)=5=c$ ($\because f\left(0\right)=b$)
$\therefore f\left(x\right)=ax+b$
On differentiating $f^{\prime }\left(x\right)=3$ and $f^{\prime \prime }\left(x\right)=0$