How do you find the range of $f (x) = {(x^{3} + 1)}^{- 1}$ $f(x)= (x^3+1)^-1$ ?

Question

How do you find the range of $f (x) = {(x^{3} + 1)}^{- 1}$ $f(x)= (x^3+1)^-1$ ?

1 Answer

Impact of this question

1556 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-09-28T15:03:15

${y ∣ y \neq 0}$ ${y|y!=0}$

Explanation:

Do find the range, get the domain of the function's inverse.

$y = {(x^{3} + 1)}^{- 1}$ $y=(x^3+1)^-1$

Flip $x$ $x$ and $y$ $y$ .

$x = {(y^{3} + 1)}^{- 1}$ $x=(y^3+1)^-1$

Now isolate $y$ $y$ .

$x = {(y^{3} + 1)}^{- 1}$ $x=(y^3+1)^-1$
$x = \frac{1}{y^{3} + 1}$ $x=1/(y^3+1)$
$(y^{3} + 1) x = (y^{3} + 1) \frac{1}{y^{3} + 1}$ $(y^3+1)x=(y^3+1)1/(y^3+1)$
$(y^{3} + 1) x = 1$ $(y^3+1)x=1$
$y^{3} x + x = 1$ $y^3x+x=1$
$y^{3} x + x - x = 1 - x$ $y^3x+x-x=1-x$
$y^{3} x = 1 - x$ $y^3x=1-x$
$(\frac{1}{x}) y^{3} x = (\frac{1}{x}) (1 - x)$ $(1/x)y^3x=(1/x)(1-x)$
$y^{3} = \frac{1 - x}{x}$ $y^3=(1-x)/x$
$\sqrt[3]{y^{3}} = \sqrt[3]{\frac{1 - x}{x}}$ $root(3)(y^3)=root(3)((1-x)/x)$
$y = \sqrt[3]{\frac{1 - x}{x}}$ $y=root(3)((1-x)/x)$
$f^{- 1} (x) = \sqrt[3]{\frac{1 - x}{x}}$ $color(blue)(f^-1(x)=root(3)((1-x)/x))$

Now looking for the domain of the inverse function, we will find that it will only be undefined at $x = 0$ $x=0$ . So its domain is ${x ∣ x \neq 0}$ ${x|x!=0}$ .

Therefore, the range of $f (x) = {(x^{3} + 1)}^{- 1}$ $f(x)=(x^3+1)^-1$ is ${y ∣ y \neq 0}$ ${y|y!=0}$ .

Science

Math

Humanities

... and beyond

How do you find the range of $f (x) = {(x^{3} + 1)}^{- 1}$ $f(x)= (x^3+1)^-1$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you find the range of f(x)=(x3+1)−1f(x)= (x^3+1)^-1?

1 Answer

Explanation:

Related questions

Impact of this question

How do you find the range of $f (x) = {(x^{3} + 1)}^{- 1}$ $f(x)= (x^3+1)^-1$ ?