Question

Prove that the function f: R->R defined by f(x) = x^2 is many one​

Answer 1

Answer:

Given, f : 1R → R; f(x) = x/1 x2 g : 1R → 1R; g(x) = 2x – 1. To show that f is neither one-one nor onto (i) f is one-one : Let x1, x2 ∈ R (domain) and f(x1) = f(x2) Read more on Sarthaks.com - https://www.sarthaks.com/217493/show-that-the-function-f-r-r-defined-by-f-x-x-x-2-1-x-r-is-neither-one-one-nor-onto?show=217505#a217505

Answer 2

Answer:

f(x) = x^2

f(x1) = f(x2)

x1^2 = x2^2

x1^2 - x2^2

(x1 - x2) (x1 + x2) = 0

Squaring x1 - x2 = 0  

x1 = x2

or, x1 = x2

or, x1 + x2 = 0

x1 = -x1

Hence, the function is many to one onto function.