Consider the functional equation f(x-y)=f(x)/f(y) . If f'(0) = p and f'(5)=q , then find f'(-5) ?

2 Answers
May 2, 2017

f'(-5)=p^2/q

Explanation:

Making x=y we have

f(0)=1

Making x=2y we have

f(y)=f(2y)/f(y) or

f(2y)=f(y)^2

or

f(ky)=f(y)^k for k in ZZ

but also with x=0

f(-y)=f(y)^-1 and

f(-ky)=f(y)^-k

Supposing now

f(y) = a^y

we have

d/(dy)f(y)=a^ylog_e a->a^0log_ea=p

or

a = e^p

and

f(y) = (e^p)^y

then

f'(5)=(e^p)^5 p=q and

f'(-5)=p/(e^p)^5

or

f'(-5) = p^2/q

May 2, 2017

p^2/q.

Explanation:

Given that, f(x-y)=f(x)/f(y).......(star).

x=y, &, (star) rArr f(0)=f(y)/f(y)=1.......(1).

x=0, y=y, & (star) rArr f(-y)=f(0)/f(y)=1/f(y).........(2).

x=x, y=-y, & (star) rArr f(x+y)=f(x)/f(-y)=f(x)f(y)

:.," by "(2), f(x+y)=f(x)f(y).............(3).

Knowing that, f'(x)=lim_(h to 0) {f(x+h)-f(x)}/h, we have, by (3),

f'(x)=lim_(h to 0){f(x)f(h)-f(x)}/h=f(x){lim_(h to 0) (f(h)-1)/h}...(ast).

Now, f'(0)=p, (ast), & (1) rArr p=lim_(h to 0) (f(h)-1)/h...(4).

Similarly, f'(5)=q, (ast), & (4) rArr q=p*f(5)......(5).

"Finally, by "(ast), &, (4), f'(-5)=p*f(-5)

=p*(1/f(5)),............[because,(2)]

=p/(q/p),................[because, (5)]

:. f'(-5)=p^2/q, as Respected Cesareo R., has derived!

Enjoy Maths.!