What is the inverse of f(x)=(x^3-2)^{1/5}-4 ?

The inverse of f(x)=(x^3-2)^{1/5}-4 is cube root of x^5+20x^4+160x^3+640x^2+1280x+1026

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inverse of f(x)=(x^3-2)^{1/5}-4

inverse

f (x) = (x^{3} - 2)^{\frac{1}{5}} - 4

Solution steps

y = (x^{3} - 2)^{\frac{1}{5}} - 4

Replace

x

with

y

x = (y^{3} - 2)^{\frac{1}{5}} - 4

Solve for

y, x = (y^{3} - 2)^{\frac{1}{5}} - 4

f (x) = \frac{3 x}{4 - 7 x}

\frac{2 x}{x ^{2} + 1}

2 - e^{x + 3}

f (x) = 5^{x} + 5^{7}

f (x) = 7 + ln (x - 2)

What is the inverse of f(x)=(x^3-2)^{1/5}-4 ?
The inverse of f(x)=(x^3-2)^{1/5}-4 is cube root of x^5+20x^4+160x^3+640x^2+1280x+1026