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

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

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

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