What is the inverse of y=sqrt(16-x^2) ?

The inverse of y=sqrt(16-x^2) is sqrt(-x^2+16),-sqrt(-x^2+16)

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inverse of y=sqrt(16-x^2)

inverse

y = 16 - x^{2}

- x^{2} + 16, - - x^{2} + 16

Solution steps

y = 16 - x^{2}

Replace

x

with

y

x = 16 - y^{2}

Solve for

y, x = 16 - y^{2}

- x^{2} + 16, - - x^{2} + 16

f (x) = \frac{5}{6 x + 1}

f (x) = - 60 (x - 3)^{2} - 50

f (x) = cos (x) + 1

f (x) = 8 x - 13

f (x) = 7 - 5 x - 2

What is the inverse of y=sqrt(16-x^2) ?
The inverse of y=sqrt(16-x^2) is sqrt(-x^2+16),-sqrt(-x^2+16)