What is the inverse of f(x)=-sqrt(9-x^2),-3<= x<0 ?

The inverse of f(x)=-sqrt(9-x^2),-3<= x<0 is sqrt(-x^2+9),-sqrt(-x^2+9)

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inverse of f(x)=-sqrt(9-x^2),-3<= x<0

inverse

f (x) = - 9 - x^{2}, - 3 \leq x < 0

- x^{2} + 9, - - x^{2} + 9

Solution steps

y = - 9 - x^{2}

Replace

x

with

y

x = - 9 - y^{2}

Solve for

y, x = - 9 - y^{2}

- x^{2} + 9, - - x^{2} + 9

What is the inverse of f(x)=-sqrt(9-x^2),-3<= x<0 ?
The inverse of f(x)=-sqrt(9-x^2),-3<= x<0 is sqrt(-x^2+9),-sqrt(-x^2+9)