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

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

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

inverse

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

- x, - - x

Solution steps

y = - x^{2}

Replace

x

with

y

x = - y^{2}

Solve for

y, x = - y^{2}

- x, - - x

y = x^{2} - 16

\frac{1}{x ^{2} - 4 x}

f (x) = 2 x^{3} + 3 x^{2} - 36 x

f (x) = x + y = - 1

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

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