What is the inverse of f(x)=10-x^2,x>= 0 ?

The inverse of f(x)=10-x^2,x>= 0 is sqrt(-x+10),-sqrt(-x+10)

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inverse of f(x)=10-x^2,x>= 0

inverse

f (x) = 10 - x^{2}, x \geq 0

- x + 10, - - x + 10

Solution steps

y = 10 - x^{2}

Replace

x

with

y

x = 10 - y^{2}

Solve for

y, x = 10 - y^{2}

- x + 10, - - x + 10

f (x) = 6 y - 3 x = - 18 x

f (x) = \frac{1}{ln ( 2 x - 1 )}

f (x) = y - 6 = 4 (x + 5)

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

(2 x - 6) (x - 4)

What is the inverse of f(x)=10-x^2,x>= 0 ?
The inverse of f(x)=10-x^2,x>= 0 is sqrt(-x+10),-sqrt(-x+10)