What is the inverse of f(x)=\sqrt[5]{x}+1 ?

The inverse of f(x)=\sqrt[5]{x}+1 is x^5-5x^4+10x^3-10x^2+5x-1

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inverse of f(x)=\sqrt[5]{x}+1

inverse

x^{5} - 5 x^{4} + 10 x^{3} - 10 x^{2} + 5 x - 1

Solution steps

Replace

x

with

y

Solve for

x^{5} - 5 x^{4} + 10 x^{3} - 10 x^{2} + 5 x - 1

f (x) = - x + 3 - 2

y = - \frac{1}{9} x - \frac{8}{9}, (1, - 4)

y = arccos (x)

sin (2.8 x + 0.9) + 0.3

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

What is the inverse of f(x)=\sqrt[5]{x}+1 ?
The inverse of f(x)=\sqrt[5]{x}+1 is x^5-5x^4+10x^3-10x^2+5x-1