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

The inverse of f(x)=(x^3-5)^{1/5}+2 is cube root of x^5-10x^4+40x^3-80x^2+80x-27

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

inverse

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

f (x) = 2 ln (x + a)

y = 45 - 0.45 x

f (x) = \frac{t ^{2} + 3 t - 4}{t ^{2} - 6 t + 5}

f (x) = e (34 - 1)

What is the inverse of f(x)=(x^3-5)^{1/5}+2 ?
The inverse of f(x)=(x^3-5)^{1/5}+2 is cube root of x^5-10x^4+40x^3-80x^2+80x-27