Answer to Question #344724 in Discrete Mathematics for Bethsheba Kiap

Question

Let f : R → R be defined by f(x) = (x3 + 1)/2

a. Prove that f is bijective

b. Determine f -1 (x) and f o f o f -1

Answer 1

a. Let $f(x_1)=f(x_2).$ It means that

$$\dfrac{x_1^3+1}{2}=\dfrac{x_2^3+1}{2}$$ $$x_1^3=x_2^3$$ $$(x_1-x_2)(x_1^2+x_1x_2+x_3^2)=0$$ $$x_1-x_2=0$$ $$x_1=x_2$$

The function $f(x)=\dfrac{x^3+1}{2}$ is bijective (one-to-one ) from $\R$ to $\R.$

b.

$$f(x)=\dfrac{x^3+1}{2}, x\in \R$$ $$y=\dfrac{x^3+1}{2}$$

Change $x$ and $y$

$$x=\dfrac{y^3+1}{2}$$

Solve for $y$

$$y^3=2x-1$$

$$y=\sqrt[3]{2x-1}$$

Then

$$f^{-1}(x)=\sqrt[3]{2x-1}$$

$$f\circ f^{-1}=\dfrac{(\sqrt[3]{2x-1})^3+1}{2}=x$$

$$f\circ f\circ f^{-1}=\dfrac{x^3+1}{2}$$