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

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

4 x + 1

f (x) = \frac{3 x + 4}{5 x + 6}

y = 5 ln (\frac{x + 3}{2})

f (x) = lo g_{2} (2 x - 8)

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)