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

The inverse of f(x)=3(x+1)^2+1 is sqrt((x-1)/3)-1,-sqrt((x-1)/3)-1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

inverse of f(x)=3(x+1)^2+1

inverse

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

\frac{x - 1}{3} - 1, - \frac{x - 1}{3} - 1

Solution steps

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

Replace

x

with

y

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

Solve for

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

\frac{x - 1}{3} - 1, - \frac{x - 1}{3} - 1

\frac{x - 3}{7} + 10

(\frac{1}{x})^{2} - 4

f (x) = \frac{9 - x ^{2}}{x + 1}

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

f (x) = \frac{x ^{2} - 16}{x + 4}

What is the inverse of f(x)=3(x+1)^2+1 ?
The inverse of f(x)=3(x+1)^2+1 is sqrt((x-1)/3)-1,-sqrt((x-1)/3)-1