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

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

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inverse of 14-x^2,x>= 0

inverse

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

∣ x - 3 ∣

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

y = \frac{2 x + 4}{1 - x}

(\frac{x ^{5}}{5} - 1)^{\frac{1}{3}}

f (x) = 3 x - 1

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