How do you find the inverse of $f (x) = \sqrt[5]{5 x + 4}$ $f(x)=root5(5x+4)$ ?

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Answer 1 · 2017-02-24T18:39:01

$f^{- 1} = \frac{x^{5} - 4}{5}$ $f^-1 = (x^5-4)/5$

Let $y = f (x) : y = \sqrt[5]{5 x + 4}$ $y = f(x): y = root(5)(5x+4)$

Swap $y$ $y$ for $x$ $x$ and $x$ $x$ for $y$ $y$ : $x = \sqrt[5]{5 y + 4}$ $x = root(5)(5y+4)$

Remember that $\sqrt[5]{} = {()}^{\frac{1}{5}}$ $root(5)( ) = ( )^(1/5)$ so $x = {(5 y + 4)}^{\frac{1}{5}}$ $x = (5y+4)^(1/5)$

Solve for $y$ $y$ .

5th power both sides: $x^{5} = {({(5 y + 4)}^{\frac{1}{5}})}^{5} = 5 y + 4$ $x^5 = ((5y+4)^(1/5))^5 = 5y+4$
Isolate $y$ $y$ : $5 y = x^{5} - 4$ $5y = x^5-4$
Simplify: $y = \frac{x^{5} - 4}{5}$ $y = (x^5-4)/5$

The inverse function $f^{- 1} = \frac{x^{5} - 4}{5}$ $f^-1 = (x^5-4)/5$

How do you find the inverse of f(x)=root5(5x+4)f(x)=5√5x+4f(x)=root5(5x+4)?