What is the inverse of f(x)=6-x^2,x>= 0 ?

The inverse of f(x)=6-x^2,x>= 0 is sqrt(-x+6),-sqrt(-x+6)

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inverse of f(x)=6-x^2,x>= 0

inverse

f (x) = 6 - x^{2}, x \geq 0

- x + 6, - - x + 6

Solution steps

y = 6 - x^{2}

Replace

x

with

y

x = 6 - y^{2}

Solve for

y, x = 6 - y^{2}

- x + 6, - - x + 6

x^{2} + 6, x \geq 0

f (x) = 10 - x^{2}

+ arcsin (x)

(2 (x + 9))^{3} - 4

f (x) = \frac{16}{25} x^{2}

What is the inverse of f(x)=6-x^2,x>= 0 ?
The inverse of f(x)=6-x^2,x>= 0 is sqrt(-x+6),-sqrt(-x+6)