What is the inverse of f(x)=x^2+4x+5,x<=-2 ?

The inverse of f(x)=x^2+4x+5,x<=-2 is -2+sqrt(x-1),-sqrt(x-1)-2

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inverse of f(x)=x^2+4x+5,x<=-2

inverse

f (x) = x^{2} + 4 x + 5, x \leq - 2

- 2 + x - 1, - x - 1 - 2

Solution steps

y = x^{2} + 4 x + 5

Replace

x

with

y

x = y^{2} + 4 y + 5

Solve for

y, x = y^{2} + 4 y + 5

- 2 + x - 1, - x - 1 - 2

- 5 x - 6

f (x) = \frac{( 2 x + 5 )}{x - 11}

tan (- \frac{2}{2})

g (x) = \frac{x + 1}{x - 1}

f (x) = \frac{1}{2} x^{2} - 8

What is the inverse of f(x)=x^2+4x+5,x<=-2 ?
The inverse of f(x)=x^2+4x+5,x<=-2 is -2+sqrt(x-1),-sqrt(x-1)-2