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

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

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

inverse

f (x) = (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

- x

f (x) = 3 x^{2} + 6 x + 5

f (x) = \frac{3}{4} - x

y = \frac{3}{4} x

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

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