What is the inverse of f(x)=-(x+1)^2-1,x<-pi/2 ?

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

inverse

f (x) = - (x + 1)^{2} - 1, x < - \frac{π}{2}

- x - 1 - 1, - - x - 1 - 1

Solution steps

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

Replace

x

with

y

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

Solve for

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

- x - 1 - 1, - - x - 1 - 1

y = ln (x)

f (x) = x^{3} + 25

- 2 x^{3} - 6

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

f (x) = \frac{( x + 8 )}{x + 6}

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