What is the inverse of f(x)=((x^7-2)/3)^{1/5} ?

The inverse of f(x)=((x^7-2)/3)^{1/5} is \sqrt[7]{3x^5+2}

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inverse of f(x)=((x^7-2)/3)^{1/5}

inverse

f (x) = (\frac{x ^{7} - 2}{3})^{\frac{1}{5}}

Solution steps

y = (\frac{x ^{7} - 2}{3})^{\frac{1}{5}}

Replace

x

with

y

x = (\frac{y ^{7} - 2}{3})^{\frac{1}{5}}

Solve for

y, x = (\frac{y ^{7} - 2}{3})^{\frac{1}{5}}

f (x) = 2 + 6 x - 2

\frac{1}{3}

f (x) = \frac{2 x + 5}{x + 2}

\frac{4 ( e ^{9} - 1 )}{3}

9 - 8 x^{3}

What is the inverse of f(x)=((x^7-2)/3)^{1/5} ?
The inverse of f(x)=((x^7-2)/3)^{1/5} is \sqrt[7]{3x^5+2}