If f:R`to`R is continous such that f(x)`-f(x/2)=(4x^(2))/3` for all `xinR` and f(0)=0, find the value of `f(3/2)`.

Explanation

We have,
`f(x)-f(x/2)=(4x^(2))/3` …(1)
`f(x/2)-f(x/4)=4/3(x/2)^(2)` (Replacing x by x/2)
`f(x/4)-f(x/8)=4/3(x/4)^(2)` [Replacing x by x/4 in (1)]
`{:(...,...,...),(...,...,...),(...,...,...):}`
Here, we have sum of infinite equations.
Since f(x) is continous and f(0)=0,
`lim_(ntooo)f(x/2^(n))=f(0)=0`
Adding above equations, we get
`f(x)=4/3x^(2)cdot1/(1-1/4)=16/9x^(2)`
`thereforef(3/2)=4`