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

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

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

inverse

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

x - 9, - x - 9

Solution steps

y = x^{2} + 9

Replace

x

with

y

x = y^{2} + 9

Solve for

y, x = y^{2} + 9

x - 9, - x - 9

f (x) = x + x + 5

f (x) = ln (9 x^{2} + 1)

y = \frac{( x - 9 )}{( x - 2 )}

f (x) = \frac{x - 2}{4 x - 16}

f (x) = \frac{1}{2 - 8 - e ^{5 t}}

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