Answer to Question #117936 in Discrete Mathematics for inv

Question

Find out if the following functions are invertible or not, If it is invertible, then find the rule of the inverse (f^(-1) (x))

1.
f:k → k^+
f(x)=x^2

2.
k^+ → k^+
f(x)=1/x

3.
f:k^+ → k^+
f(x)=x^2

Answer 1

A function is invertible if the function is surjective and injective.

(More infomartion: https://en.wikipedia.org/wiki/Inverse_function)

Definition. Let $f$  be a function whose domain is a set $X$ . The function $f$  is said to be injective provided that for all $a$  and $b$  in $X$ , whenever $f(a)=f(b)$ , then $a=b$ ; that is, $f(a)=f(b)$  implies $a=b$ . Equivalently, if $a\neq b$ , then $f(a)\neq f(b)$ .

Symbolically,

$$\displaystyle \forall a,b\in X,\;\;f(a)=f(b)\Rightarrow a=b$$

which is logically equivalent to the contrapositive,

$$\displaystyle \forall a,b\in X,\;\;f(a)\neq f(b)\Rightarrow a\neq b$$

(More infomation: https://en.wikipedia.org/wiki/Injective_function)

Definition. A surjective function is a function whose image is equal to its codomain. Equivalently, a function $f$  with domain $X$  and codomain $Y$  is surjective, if for every $y$  in $Y$ , there exists at least one $x$  in $X$  with $\displaystyle f(x)=y$ .

Symbolically,

If $\displaystyle f\colon X\rightarrow Y$ , then $\displaystyle f$  is said to be surjective if

$$\displaystyle \forall y\in Y,\,\exists x\in X,\;\;f(x)=y$$

(More information: https://en.wikipedia.org/wiki/Surjective_function)

Hint: For all these questions, I will consider that $k\equiv\reals$ .

In our case,

(1) $f : \reals\rightarrow\reals^+;\quad f(x)=x^2$

The function $f(x)=x^2$ is not injective in the domain $D(f)=\reals$ , since

$$(-3)^2=9=3^2,\quad\text{but}\quad -3\neq3$$

Conclusion.

$$\boxed{f : \reals\rightarrow\reals^+;\quad f(x)=x^2-\text{no inverse function}}$$

(2) $f : \reals^+\rightarrow\reals^+;\quad f(x)=1/x$

This function is surjective and injective, therefore the function has an inverse function.

$$y=\frac{1}{x}\longrightarrow x=\frac{1}{y}\longrightarrow\\[0.3cm] \boxed{f(x)=\frac{1}{x}\longrightarrow f^{-1}(x)=\frac{1}{x}}$$

Verification,

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

(3) $f : \reals^+\rightarrow\reals^+;\quad f(x)=x^2$

This function is surjective and injective, therefore the function has an inverse function.

$$y=x^2\longrightarrow x=\sqrt{y}\longrightarrow\\[0.3cm] \boxed{f(x)=x^2\longrightarrow f^{-1}(x)=\sqrt{x}}$$

Verification,

$$f\left(f^{-1}(x)\right)=\left(\sqrt{x}\right)^2=x$$

ANSWER

(1) $f : \reals\rightarrow\reals^+;\quad f(x)=x^2$

$$f : \reals\rightarrow\reals^+;\quad f(x)=x^2-\text{no inverse function}$$

(2) $f : \reals^+\rightarrow\reals^+;\quad f(x)=1/x$

$$f(x)=\frac{1}{x}\longrightarrow f^{-1}(x)=\frac{1}{x}$$

(3) $f : \reals^+\rightarrow\reals^+;\quad f(x)=x^2$

$$f(x)=x^2\longrightarrow f^{-1}(x)=\sqrt{x}$$