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

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

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inverse of f(x)=3-x^2,x>= 0

inverse

f (x) = 3 - x^{2}, x \geq 0

- x + 3, - - x + 3

Solution steps

y = 3 - x^{2}

Replace

x

with

y

x = 3 - y^{2}

Solve for

y, x = 3 - y^{2}

- x + 3, - - x + 3

2 lo g_{10} (x - 3) + 2

f (x) = ln (1 + x)

y = \frac{b ^{x} + b ^{- x}}{2}

10 - 2 p^{\frac{1}{2}}

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