What is the inverse of f(x)=-x^2+1,x>= 0 ?

The inverse of f(x)=-x^2+1,x>= 0 is sqrt(-x+1),-sqrt(-x+1)

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inverse of f(x)=-x^2+1,x>= 0

inverse

f (x) = - x^{2} + 1, x \geq 0

- x + 1, - - x + 1

Solution steps

y = - x^{2} + 1

Replace

x

with

y

x = - y^{2} + 1

Solve for

y, x = - y^{2} + 1

- x + 1, - - x + 1

f (x) = - 2 (x + 1)^{3}

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

y = 2^{x - 1} + 5

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

\frac{8}{x + 3}

What is the inverse of f(x)=-x^2+1,x>= 0 ?
The inverse of f(x)=-x^2+1,x>= 0 is sqrt(-x+1),-sqrt(-x+1)