Question

Functions returns what when x is 5(f(5))?

Answer 1

Answer:

Setting y=0 into that equation we deduce:

f(x)=f(x+0)=f(x)*f(0)-f(x*0)+1=f(0)f(x)-f(0)+1

that is

(1-f(0))*(f(x)-1)=0

Hence, either

f(0)=1

or

f(x)=1 for all values of x.

The second case f(x)=1 for all values of x rejected since it contradicts f(1) = 2.

In the first case we set y=1 into the given equation and since f(1)=2 we get

f(x+1)=f(x)*f(1)-f(x*1)+1=(f(1)-1)*f(x)+1=(2–1)*f(x)+1=f(x)+1

from which we can easily deduce by induction, that for every natural number n:

f(x+n)=f((x+n-1)+1)=f(x+n-1)+1=…=f(x)+n

and in particular, for x=0 it readily follows that

f(0+n)=f(n)=f(0)+n=1+n ===> f(5)=6.

Explanation:

hope u have been understood