What is the inverse of f(x)=sqrt(16-x^2),0<= x<= 4 ?

The inverse of f(x)=sqrt(16-x^2),0<= x<= 4 is sqrt(-x^2+16),-sqrt(-x^2+16)

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inverse of f(x)=sqrt(16-x^2),0<= x<= 4

inverse

f (x) = 16 - x^{2}, 0 \leq x \leq 4

- 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

lo g_{3} (x + 1)

\frac{5 x}{x + 2}

f (x) = ln (2 + x)

f (x) = \frac{3 x}{2} + 4

f (x) = \frac{4 x ^{2}}{7}

What is the inverse of f(x)=sqrt(16-x^2),0<= x<= 4 ?
The inverse of f(x)=sqrt(16-x^2),0<= x<= 4 is sqrt(-x^2+16),-sqrt(-x^2+16)