Answer to Question #174750 in Algebra for Hetisani Sewela

Question

Consider the following two functions:

1. f : R → R defined by f(x) = 4x − 15.

2. g : R → R defined by f(x) = 15x³.

Prove that both f and g are one-to-one correspondences.

Let f : A → B be a one-to-one correspondence. Then to each b ∈ B there corresponds a unique a ∈ A such that f(a) = b. We define f^-1: B → A by

f^-1(b) = the unique a such that f(a) = b.

Answer 1

$\text{Let a,b}\in R$

$f(a)=4a-15$

$f(b) = 4b-15$

$\text{Assuming f(a) = f(b)}$

$\implies 4a-15=4b-15$

$\implies a=b$

$\text{Hence f is one to one correspondence since f(a) = f(b) implies a = b }$

$\text{Also let } g(a)=15a^3 \text{ and }g(b)=15b^3$

$\text{Assuming g(a) = g(b)}$

$\implies 15a^3=15b^3$

$\implies a=b$

$\text{Hence g is one to one correspondence since g(a) = g(b) implies a = b }$