Question

A function f:R→R defined by $f(x)=\frac{3x}{5}+2$, x ∈ R. Show that f is one-one and onto. Hence find f⁻¹.

Answer 1

A function f : R ----> R defined by $f(x)=\frac{3}{5}x+2$, $x\in R$

let's take $x_1$ and $x_2$ are two numbers in domain of given function such that, $f(x_1)=f(x_2)$

or, $\frac{3}{5}x_1+2=\frac{3}{5}x_2+2$

or, $\frac{3}{5}x_1=\frac{3}{5}x_2$

or, $x_1=x_2$

hence, f is one - one function.

we know, every function is an onto function only when co - domain = range.

here function ,f(x) = 3/5 x + 2 is a linear polynomial function. we know, every polynomial function is defined in all real numbers and range of polynomial function belongs to all real numbers.

given, co - domain belongs to R
so, range belongs to R

so, f is onto function.

now, f(x) = 3/5 x + 2

or, y = 3/5 x + 2

or, y - 2 = 3/5 x

or, x = 5(y - 2)/3

hence, $f^{-1}=\frac{5(x-2)}{3}$