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

The inverse of f(x)=(7x^3+2)^{1/5} is cube root of (x^5-2)/7

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

inverse

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

7 sin (4 (t + 7)) - 8

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

f (x) = 2 x - ln (2 x)

- 1 - x

14 (x - 4) (x + 10)

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