Question

Question: show that function $\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathit{a}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathit{b}$from R to R is invertible, where a and b are constants, with$\mathit{a}\mathbf{\ne }\mathbf{0}$, and find the inverse of f.

Short answer

Answer:

The inverse function is $f^{-1}\left(x\right)=\frac{x-b}{a}=\frac{1}{a}x-\frac{b}{a}$

Step-by-step solution

Step 1:

The function f is one-to-one if and only if $\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\mathit{f}\mathbf{\left(}\mathit{b}\mathbf{\right)}$ implies that for all a and b in the domain.

The function f is onto if and only if for every element $\mathit{b}\mathbf{\in }\mathit{B}$there exist an element$\mathit{a}\mathbf{\in }\mathit{A}$ such that$\mathit{f}\mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{=}\mathit{b}$

F is one-to one correspondence if and only if f is one-to-one and onto.

If f is one-to-one correspondence, then f is invertible.

Step 2:

Given:

$$\begin{gathered} f:\mathrm{ℝ}\to \mathrm{ℝ} \\ f\left(x\right)=ax+b \\ a\ne 0 \end{gathered}$$

ONE-TO-ONE

Let us assume $f\left(c\right)=f\left(d\right).$since$f\left(x\right)=ax+b$

$ac+b=ad+b$

Subtract b from each side of the previous equation:

$ac=ad$

Divide each side of the previous equation by a

$c=d$

Thus $f\left(c\right)=f\left(d\right)$implies and$c=d$ thus by the definition of one-to-one, f is one-to-one.

Step 3:

ONTO

Let $y\in R$ . then$\frac{y-b}{a}\in R$ with the property:

$f\left(\frac{y-b}{a}\right)=a.\frac{y-b}{a}+b=(y-b)+b=y$

By the definition of onto, f is then onto.

INVERTIBLE

Since f is one-to-one and onto, f is one-to-one correspondence and thus f is also invertible.

Step 4:

$f\left(x\right)=ax+b_{}$

Replace f(x) by y:

$y=ax+b$

To determine the inverse, we interchanging x and y:

$x=ay+b$

Next we will solve the equation to y. start by subtracting b from each side of the previous equation:

$x-b=ay$

Divide each side of the previous equation by a

$\frac{x-b}{a}=y$

Finally replace y by $f^{-1}\left(x\right)$and interchange the sides of the previous equation:

$f^{-1}\left(x\right)=\frac{x-b}{a}=\frac{1}{a}x-\frac{b}{a}$

Hence, the solution is

$f^{-1}\left(x\right)=\frac{x-b}{a}=\frac{1}{a}x-\frac{b}{a}$