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Intermediate Algebra

OPENSTAX

$Math Studyset 91影视 Explanations$ Math

OER 2017 Edition 路 1346 pages

ISBN: 9780998625720

Chapter 10: Q. 10.17 (page 1001)

Find the inverse of the function , $f (x) = \sqrt[5]{3 x - 2}$

Short Answer

Expert verified

The inverse of, $f (x) = \sqrt[5]{3 x - 2}$ is

$f^{- 1} (x) = \frac{x^{5} + 2}{3}$

Step by step solution

01

Given information

Given the function, $f (x) = \sqrt[5]{3 x - 2}$

02

Replace f(x) with yand then x,y

Replacing, we get

$f (x) = \sqrt[5]{3 x - 2} y = \sqrt[5]{3 x - 2} x = \sqrt[5]{3 y - 2} x^{5} = 3 y - 2$

03

Solve further and calculate the inverse

Further solving, we get

$x^{5} = 3 y - 2 x^{5} + 2 = 3 y y = \frac{x^{5} + 2}{3} f^{- 1} (x) = \frac{x^{5} + 2}{3}$

04

Verify f(f-1(x))=x

Substituting, we get

$f (\frac{x^{5} + 2}{3}) = \sqrt[5]{3 (\frac{x^{5} + 2}{3}) - 2} f (\frac{x^{5} + 2}{3}) = \sqrt[5]{x^{5} + 2 - 2} f (\frac{x^{5} + 2}{3}) = \sqrt[5]{x^{5}} f (\frac{x^{5} + 2}{3}) = x$

05

Verify f-1(f(x))=x

Substituting, we get

$f^{- 1} (\sqrt[5]{3 x - 2}) = \sqrt[5]{3 (\frac{x^{5} + 2}{3}) - 2} f^{- 1} (\sqrt[5]{3 x - 2}) = \sqrt[5]{x^{5} + 2 - 2} f^{- 1} (\sqrt[5]{3 x - 2}) = \sqrt[5]{x^{5}} f^{- 1} (\sqrt[5]{3 x - 2}) = x$

So these are inverses

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