What is the inverse of f(x)=8-x^2,x>= 0 ?

The inverse of f(x)=8-x^2,x>= 0 is sqrt(-x+8),-sqrt(-x+8)

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inverse of f(x)=8-x^2,x>= 0

inverse

f (x) = 8 - x^{2}, x \geq 0

- x + 8, - - x + 8

Solution steps

y = 8 - x^{2}

Replace

x

with

y

x = 8 - y^{2}

Solve for

y, x = 8 - y^{2}

- x + 8, - - x + 8

\frac{x ^{2} - 2 x - 15}{x}

f (x) = x^{2} - 12

f (x) = \frac{7 x ^{2} - 2 x}{15 x + 5}

y = 6 (7 x + 150)

y = - e^{- x} + 3 - x^{3} + 2 x

What is the inverse of f(x)=8-x^2,x>= 0 ?
The inverse of f(x)=8-x^2,x>= 0 is sqrt(-x+8),-sqrt(-x+8)