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

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

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

inverse

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

- x + 4, - - x + 4

Solution steps

y = 4 - x^{2}

Replace

x

with

y

x = 4 - y^{2}

Solve for

y, x = 4 - y^{2}

- x + 4, - - x + 4

in v erse f (x) = 3 - 2 y^{3}

d o main f (x) = 6 x + 4 + 7 x + 8

in v erse \frac{2 x}{x + 3}

d o main f (x) = x - 5

a sy m pt o t es f (x) = 2^{x} + 2

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