What is the inverse of f(x)=14-x^2,x>= 0 ?

The inverse of f(x)=14-x^2,x>= 0 is sqrt(-x+14),-sqrt(-x+14)

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inverse of f(x)=14-x^2,x>= 0

inverse

f (x) = 14 - x^{2}, x \geq 0

- x + 14, - - x + 14

Solution steps

y = 14 - x^{2}

Replace

x

with

y

x = 14 - y^{2}

Solve for

y, x = 14 - y^{2}

- x + 14, - - x + 14

f (x) = 4 (\frac{x ^{\frac{1}{7}}}{5} - 4)

f (x) = x^{2} + 4 x + 2

2 e^{3 x}

\frac{1}{x - 6}

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

What is the inverse of f(x)=14-x^2,x>= 0 ?
The inverse of f(x)=14-x^2,x>= 0 is sqrt(-x+14),-sqrt(-x+14)