0

Problem: A function $f: \mathbb R \to \mathbb R$ satisfies equation $f(x+y)=f(x)f(y)$ for all $x,y\in \mathbb R$ and $f(x)\neq 0$ for all $x \in \mathbb R$.
If $f'(0)=2$, find $f'(x)$ in terms of $f(x)$

Solution:
$f(x+y)=f(x)f(y)$
Differentiating partially w.r.t $x$ we get
$f'(x+y)=f'(x)f(y)$
At x=0
$f'(y)=f'(0)f(y)$
$f'(y)= 2f(y)$
$f'(x)= 2f(x)$

Can we solve question like above ?

I mean,$f'(x)$ represent ordinary diff. while I used it for partial diff.

Community
  • 1
rst
  • 2,031

1 Answers1

5

More technically, you can't do like that(why?)

Since you are not given that $f$ is differentiable(except at $0$), therefore you can't take derivatives on both sides. So, first you need to check if $f$ is differentiable and in that process, you'll get $f'(x)$( if $f$ comes out to be differentiable on some non-empty set of $\Bbb R$)

Note, $f(0+0)=f(0)f(0)\implies f(0)=0,1$ but $f(x)\neq 0$ for any $x\in \Bbb R\implies f(0)=1$

Then, $1=f(0)=f(x-x)=f(x).f(-x)\implies f(-x)=\frac{1}{f(x)}$

Now, $$R.H.D.=\lim_{h\to 0^+}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0^+}\frac{f(x)f(h)-f(x)}{h}$$ $$=f(x)\lim_{h\to 0^+}\frac{f(h)-1}{h}\tag{1}$$

similarly, $$L.H.D.=\lim_{h\to 0^+}\frac{f(x)-f(x-h)}{h}=\lim_{h\to 0^+}\frac{f(x)-f(x)f(-h)}{h}$$ $$=f(x)\lim_{h\to 0^+}\frac{1-\frac{1}{f(h)}}{h}=f(x)\left(\lim_{h\to 0^+}\frac{f(h)-1}{h}\right)\left(\lim_{h\to 0^+}\frac{1}{f(h)}\right)=f(x)\lim_{h\to 0^+}\frac{f(h)-1}{h}\tag{2}$$

From $(1)$ and $(2)$, $L.H.D.=R.H.D.\implies f$ is differential.

Then, $$f'(x)=f(x)\lim_{h\to 0^+}\frac{f(h)-1}{h}$$

Since, $2=f'(0)=\lim_{h\to 0^+}\frac{f(h)-f(0)}{h}=\lim_{h\to 0^+}\frac{f(h)-1}{h}$

Therefore, $$f'(x)=2f(x)\tag{QED}$$

Aang
  • 14,672