Question

Let $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x,yϵR$ and suppose that $f$ is differentiable at 0 and $f^{\prime }\left(0\right)=4$. If $f\left(x_0\right)=8$ then $f^{\prime }\left(x_0\right)$ is equal to

Answer 1

put $y=0$. Then $f\left(x\right)=f\left(x\right)f\left(0\right),i.e.,f\left(x\right)\left[f\right(0)-1]=0.$ So $f\left(x\right)=0orf\left(0\right)=1.$
If$f\left(x\right)=0$ for all x, then f is clearly differentiable.
therefore,f is a nonzero function, so that $f\left(0\right)=1.$

Since f is differentiable at $x=0,$ we have
$f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f(0+h)-f\left(0\right)}{h}$

$=\underset{h\to 0}{lim}\frac{f\left(0\right)f\left(h\right)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}$
Let $x_0ϵR$. Then,
$\Rightarrow \underset{h\to 0}{lim}\frac{f(x_0+h)-f\left(x_0\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(x_0\right)f\left(h\right)-f\left(x_0\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(x_0\right)[f\left(h\right)-1]}{h}=f\left(x_0\right)\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}=f\left(x_0\right)f^{\prime }\left(0\right)=32$