What is the inverse of f(x)=-x^2+5,x>= 0 ?

The inverse of f(x)=-x^2+5,x>= 0 is sqrt(-x+5),-sqrt(-x+5)

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inverse of f(x)=-x^2+5,x>= 0

inverse

f (x) = - x^{2} + 5, x \geq 0

- x + 5, - - x + 5

Solution steps

y = - x^{2} + 5

Replace

x

with

y

x = - y^{2} + 5

Solve for

y, x = - y^{2} + 5

- x + 5, - - x + 5

(x - \frac{1}{4})^{2} + 2

\frac{9 x - 4}{- 3 x + 1}

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

\frac{\frac{4}{x} - 12}{x - \frac{1}{9}}

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

What is the inverse of f(x)=-x^2+5,x>= 0 ?
The inverse of f(x)=-x^2+5,x>= 0 is sqrt(-x+5),-sqrt(-x+5)