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 \leq 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

3 x^{3} - 2

f (x) = \frac{x}{3} - 7

- x^{2}

f (x) = 2 x - 8

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)