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

3 x + 4 y = 5, (3, - 2)

lo g_{3} (x^{2} - 1)

f (x) = 3 2 x - 1 + 15

x^{3} + 3 x + 6

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

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)