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

The inverse of f(x)=(x^3+5)^{1/5}-5 is cube root of x^5+25x^4+250x^3+1250x^2+3125x+3120

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+5)^{1/5}-5

inverse

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

x + 4 - 1

- \frac{42475}{12} x^{2} + 405 x - 70

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

\frac{2 x ^{2} + 2 x}{- 3 x ^{2} - 18 x - 15}

f (x) = e^{8 x - 6}

What is the inverse of f(x)=(x^3+5)^{1/5}-5 ?
The inverse of f(x)=(x^3+5)^{1/5}-5 is cube root of x^5+25x^4+250x^3+1250x^2+3125x+3120