Question

Given that f(x) =


`\sqrt[3]{x} \: \: + 5`
is one-to- one, find each of the following:
a) Find

`{f}^( - 1) (x)`
, the equation of the inverse of f(x)

`{f}^( - 1) (x)`
=

b) Find the coordinates of f (x )and

`{f}^( - 1) (x)`

c) Find the domain and range of both f(x) and

`{f}^( - 1) (x)`

Answer 1

`f(x)=\sqrt[3]{x} +5\\\\\implies y = √(3){x}+5`

a) Since the function is given to be one - one so the inverse of the function exist. Now f(x) maps x to y so the inverse of f(x) maps y to x

To find inverse, first interchange the roles of x and y :

`\implies x = \sqrt[3]{y}+5`

Now, solve for y :

`x = \sqrt[3]{y}+5\\\\\implies \sqrt[3]{y}=x-5\\\\\text{Now, cubing both the sides. We get,}\\\\\implies y=(x-5)^3\\\\\implies y=x^3-15\cdot x^2+75\cdot x-125\\\\\implies\bf f^(-1)(x)=x^3-15\cdot x^2+75\cdot x-125`

b) To find coordinates of f(x) :

`f(x)=\sqrt[3]{x} +5`

First take y = 0 then take x = 0

⇒ x-coordinates : (-125,0) and y-coordinates : (0,5)

To find coordinates of inverse function of x :

`y=x^3-15\cdot x^2+75\cdot x-125`

First take y = 0 then take x = 0

⇒ x - coordinates : (5,0) and y - coordinates : (0,-125)

c) f(x) is defined for every real number.

⇒ Domain : -∞ < x < ∞

and Range : -∞ < f(x) < ∞

And inverse function of x is also defined for every real number :

⇒ Domain : -∞ < x < ∞

`\text{and Range : }-\infty < f^(-1)(x) < \infty`