What is the inverse of f(x)=1+x^2,x>= 0 ?

The inverse of f(x)=1+x^2,x>= 0 is sqrt(x-1),-sqrt(x-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)=1+x^2,x>= 0

inverse

f (x) = 1 + x^{2}, x \geq 0

x - 1, - x - 1

Solution steps

y = 1 + x^{2}

Replace

x

with

y

x = 1 + y^{2}

Solve for

y, x = 1 + y^{2}

x - 1, - x - 1

f (x) = 8 - 7 x^{3}

f (x) = - lo g_{9} (x - 4) - 6

f (x) = \frac{7 - 3 x}{2 x + 3}

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

\frac{1}{2} e^{2 x} + 2

What is the inverse of f(x)=1+x^2,x>= 0 ?
The inverse of f(x)=1+x^2,x>= 0 is sqrt(x-1),-sqrt(x-1)