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

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

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inverse of f(x)=9-x^2,x>= 0

inverse

f (x) = 9 - x^{2}, x \geq 0

- x + 9, - - x + 9

Solution steps

y = 9 - x^{2}

Replace

x

with

y

x = 9 - y^{2}

Solve for

y, x = 9 - y^{2}

- x + 9, - - x + 9

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

9 x^{2} + 4 y^{2} = 1

f (x) = x^{2} + x + 2

f (x) = 2 x - 3 x^{\frac{2}{3}}

\frac{( 4 x + 9 )}{6 x + 3}

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