How do you find the inverse of $y = x^{3} + 5$ $y=x^3 +5$ ?

Question

10738 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-04T00:48:28

Reformulate the equation with $x$ $x$ isolated on one side to find inverse function:

$f^{- 1} (y) = \sqrt[3]{y - 5}$ $f^(-1)(y) = root(3)(y-5)$

Let $f (x) = x^{3} + 5$ $f(x) = x^3+5$

Let $y = f (x) = x^{3} + 5$ $y = f(x) = x^3+5$

Subtract $5$ $5$ from both ends to get:

$y - 5 = x^{3}$ $y - 5 = x^3$

Take the cube root of both sides and transpose to get:

$x = \sqrt[3]{y - 5}$ $x = root(3)(y - 5)$

Hence:

$f^{- 1} (y) = \sqrt[3]{y - 5}$ $f^(-1)(y) = root(3)(y-5)$

How do you find the inverse of y=x3+5y=x^3 +5?