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

The inverse of f(x)=(x^3+1)^{1/5}+2 is cube root of x^5-10x^4+40x^3-80x^2+80x-33

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse

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

3 x^{5} - 10 x^{4} + 40 x^{3} - 80 x^{2} + 80 x - 33

Solution steps

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

Replace

x

with

y

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

Solve for

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

3 x^{5} - 10 x^{4} + 40 x^{3} - 80 x^{2} + 80 x - 33

f (x) = \frac{4 x - 1}{7 x + 5}

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

lo g_{x} (2)

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

\frac{6}{8 - x}

What is the inverse of f(x)=(x^3+1)^{1/5}+2 ?
The inverse of f(x)=(x^3+1)^{1/5}+2 is cube root of x^5-10x^4+40x^3-80x^2+80x-33