Question

Let $f(x+y)=f\left(x\right)f\left(y\right)$ for all $x$ and $y$. Suppose $f\left(5\right)=2$ and $f^{\prime }\left(0\right)=3.$ Find $f^{\prime }\left(5\right).$

Answer 1

We are given $f(x+y)=f\left(x\right)f\left(y\right)...\left(1\right)$ for all x and y.

In the equation (1),we put $x=5,y=0.$ Then $f\left(5\right)=f\left(5\right)f\left(0\right).$

This gives $f\left(0\right)=1[\because f\left(5\right)=2\ne 0].$

Now $f^{\prime }\left(5\right)=\underset{h\to 0}{lim}\frac{f(5+h)-f\left(5\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(5\right)f\left(h\right)-f\left(5\right)}{h}$ by (1)

$\Rightarrow f^{\prime }\left(5\right)=f\left(5\right)\underset{h\to 0}{lim}\frac{f\left(h\right)-1}{h}=f\left(5\right)\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}[\because f\left(0\right)=1]$

$f^{\prime }\left(5\right)=f\left(5\right).f^{\prime }\left(0\right)=2\times 3=6$