What is the inverse of f(x)=11-x^2,x>= 0 ?

The inverse of f(x)=11-x^2,x>= 0 is sqrt(-x+11),-sqrt(-x+11)

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inverse of f(x)=11-x^2,x>= 0

inverse

f (x) = 11 - x^{2}, x \geq 0

- x + 11, - - x + 11

Solution steps

y = 11 - x^{2}

Replace

x

with

y

x = 11 - y^{2}

Solve for

y, x = 11 - y^{2}

- x + 11, - - x + 11

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

f (x) = (x - 11)^{2}

x^{3} - 3 x^{2} + 3 x - 1

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

g (x) = x^{3} - 5

What is the inverse of f(x)=11-x^2,x>= 0 ?
The inverse of f(x)=11-x^2,x>= 0 is sqrt(-x+11),-sqrt(-x+11)