What is the inverse of f(x)=x^2-5,x<= 0 ?

The inverse of f(x)=x^2-5,x<= 0 is sqrt(x+5),-sqrt(x+5)

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inverse of f(x)=x^2-5,x<= 0

inverse

f (x) = x^{2} - 5, x \leq 0

x + 5, - x + 5

Solution steps

y = x^{2} - 5

Replace

x

with

y

x = y^{2} - 5

Solve for

y, x = y^{2} - 5

x + 5, - x + 5

3 sin (\frac{x}{π} + \frac{1}{2}) - 2

lo g_{3} (x + 2)

2 x - 3

f (x) = - 4^{- x - 3} + 5

f (x) = - 8

What is the inverse of f(x)=x^2-5,x<= 0 ?
The inverse of f(x)=x^2-5,x<= 0 is sqrt(x+5),-sqrt(x+5)