Question

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

Short 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 $f\left(a\right)=f\left(b\right)$ implies that for all a and b in the domain.

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

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:

$f:\mathrm{ℝ}\to \mathrm{ℝ}$

$\begin{array}{r}f\left(x\right)=ax+b\\ a\ne 0\end{array}$

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 c = d and 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:

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

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:

data-custom-editor="chemistry" $\mathrm{x}-\mathrm{b}=\mathrm{ay}$

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