Answer to Question #343475 in Discrete Mathematics for Bethsheba Kiap

Question

Two functions f : R → R and g : R → R are defined by f(x) = 5x3 + 1 and g(x) = 2x − 3 for all x ∈ R.

Determine the inverse of (f -1 ◦ g) and (g ◦ f )(2) and ( f ◦ g)(2).

Answer 1

Let $f(x_1)=f(x_2).$ It means that

$$5x_1^3+1=5x_2^3+1$$ $$x_1^3=x_2^3$$ $$(x_1-x_2)(x_1^2+x_1x_2+x_3^2)=0$$ $$x_1-x_2=0$$ $$x_1=x_2$$

The function $f(x)=5x^3+1$ is bijective (one-to-one ) from $\R$ to $\R.$

$$f(x)=5x^3+1, x\in \R$$ $$y=5x^3+1$$

Change $x$ and $y$

$$x=5y^3+1$$

Solve for $y$

$$y^3=\dfrac{x-1}{5}$$

$$y=\sqrt[3]{\dfrac{x-1}{5}}$$

Then

$$f^{-1}(x)=\sqrt[3]{\dfrac{x-1}{5}}, x\in \R$$

a)

$$(f^{-1}\circ g)=\sqrt[3]{\dfrac{2x-3-1}{5}}=\sqrt[3]{\dfrac{2x-4}{5}}, x\in \R$$

$$y=\sqrt[3]{\dfrac{2x-4}{5}}$$

Change $x$ and $y$

$$x=\sqrt[3]{\dfrac{2y-4}{5}}$$

Solve for $y$

$$2y-4=5x^3$$

$$y=\dfrac{5}{2}x^3+2$$

Then

$$(f^{-1}\circ g)^{-1}=\dfrac{5}{2}x^3+2, x\in \R$$

b)

$$(g\circ f)(x)=2(5x^3+1)-3=10x^3-1$$

$$(g\circ f)(2)=10(2)^3-1=79$$

c)

$$(f\circ g)(x)=5(2x-3)^3+1$$

$$(f\circ g)(2)=5(2(2)-3)^3+1=6$$