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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of f(x)=\sqrt[5]{x-3}+1

inverse

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

Solution steps

Replace

x

with

y

Solve for

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

x + 4

f (x) = (x + 8)^{3}

f (x) = (10 - 10 x)^{\frac{5}{2}}

f (x) = \frac{4}{x} + 6

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

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