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

The inverse of f(x)=(x^3-2)^{1/5}+3 is cube root of x^5-15x^4+90x^3-270x^2+405x-241

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

inverse

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

f (x) = sin (\frac{1 - 2 x}{x})

\frac{e ^{x + 2}}{e ^{x + 2} - 4}

f (x) = \frac{3 x}{2} - 12

y = x (- x + 8)

\frac{28}{s ^{6}}

What is the inverse of f(x)=(x^3-2)^{1/5}+3 ?
The inverse of f(x)=(x^3-2)^{1/5}+3 is cube root of x^5-15x^4+90x^3-270x^2+405x-241