$\text{ If }f\left(\frac{x+y}{2}\right)=\frac{f\left(x\right)+f\left(y\right)}{2}\text{ for all real }x\text{ and }y\text{ and }{f}^1\left(0\right)\text{ exists and equals to }-1\text{ and }f\left(0\right)=1\text{ , }$then $f^1$(x)=

$f\left(\frac{x+y}{2}\right)=\frac{f\left(x\right)+f\left(y\right)}{2}\mathrm{....}\left(1\right)$

Put y=0, f(0)=1

$f\left(\frac{x}{2}\right)=\frac{1}{2}\left(f\left(x\right)+1\right)\Rightarrow f\left(x\right)=2f\left(\frac{x}{2}\right)-1\to \left(2\right)$

$f\text{'}\left(x\right)=\underset{h\to 0}{\mathrm{l}t}\frac{f\left(x+h\right)-f\left(x\right)}{h}$

$=\underset{h\to 0}{\mathrm{l}t}\frac{f\left(\frac{2x+2h}{2}\right)-f\left(x\right)}{h}$

$=\underset{h\to 0}{\mathrm{l}t}\frac{\frac{f\left(2x\right)+f\left(2h\right)}{2}-f\left(x\right)}{h}$

$=\underset{h\to 0}{\mathrm{l}t}\frac{f\left(2x\right)+f\left(2h\right)-2f\left(x\right)}{2h}$

$=\underset{h\to 0}{\mathrm{l}t}\frac{2f\left(x\right)-1+f\left(2h\right)-2f\left(x\right)}{2h}$

$=\underset{h\to 0}{\mathrm{l}t}\frac{f\left(2h\right)-1}{2h}=f\text{'}\left(0\right)$

$f\text{'}\left(x\right)=-1$