**MATH-TRICK-6**

*This is on Functional Equations explained through some solved problems.*

**Result:- If f is a continuous function that satisfies f(x+y)=f(x)+f(y) for all x,y, then f(x)=xf(1).**

Proof:- (Hints)

We know that if f is a function satisfying f(x+y)=f(x)+f(y) for all x,y then f(x)=kx ,where k is a constant.

[To show this, Check the differentiability first.Then

f'(x)=Lt{f(x+h)-f}/h ,where h->0.

f'(x)=Lt {f(h)/h} where h->0

= f'(0)=k(say).

Then integrate both sides, you will get f(x)=kx+c ------(1)

Put x=y=0 in the given equation,then f(0)=0.

From (1), c=0, i.e. f(x)=kx. ]

Here f(1)=k ,so f(x)=kx=xf(1).

**Problem:- [ISI Sample paper] Let g be a continuous function with g(1)=1 such that**

**g(x+y)=5g(x)g(y)**

**for all x,y. Find g(x).**

Sol:- g(x+y)=5g(x)g(y)

5g(x+y)=5g(x)5g(y), taking log bothsides

=> log[5g(x+y)]=log[5g(x)] + log[5g(y)]

=> f(x+y)=f(x)+f(y), taking f(x+y)=log[5g(x+y)]

=> f(x)= xf(1) [By the above result]

So, log[5g(x)]=xlog[5g(1)]

=> 5g(x) = 5^x ,since g(1)=1 (given)

=> g(x) = 5^{x-1} .