What is the inverse of f(x)=-sqrt(4-x^2),0<= x<2 ?

The inverse of f(x)=-sqrt(4-x^2),0<= x<2 is sqrt(-x^2+4),-sqrt(-x^2+4)

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inverse of f(x)=-sqrt(4-x^2),0<= x<2

inverse

f (x) = - 4 - x^{2}, 0 \leq x < 2

- x^{2} + 4, - - x^{2} + 4

Solution steps

y = - 4 - x^{2}

Replace

x

with

y

x = - 4 - y^{2}

Solve for

y, x = - 4 - y^{2}

- x^{2} + 4, - - x^{2} + 4

f (x) = 3 x + 7

\frac{81 x ^{4}}{16}

f (x) = x^{2} + 16 + 2 x

y = \frac{8}{4 - 2 x}

f (x) = \frac{2}{5} x + \frac{1}{3}

What is the inverse of f(x)=-sqrt(4-x^2),0<= x<2 ?
The inverse of f(x)=-sqrt(4-x^2),0<= x<2 is sqrt(-x^2+4),-sqrt(-x^2+4)