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

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

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inverse of f(x)=x^2-1,x<= 0

inverse

f (x) = x^{2} - 1, x \leq 0

x + 1, - x + 1

Solution steps

y = x^{2} - 1

Replace

x

with

y

x = y^{2} - 1

Solve for

y, x = y^{2} - 1

x + 1, - x + 1

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

f (x) = \frac{1}{x} + 1

3 x^{2} - 2

f (x) = \frac{1}{x - 3} + 4

f (x) = - 10 6 - 11 x + 10

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