What is the inverse of f(x)=x^2+8,x>= 0 ?

The inverse of f(x)=x^2+8,x>= 0 is sqrt(x-8),-sqrt(x-8)

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inverse of f(x)=x^2+8,x>= 0

inverse

f (x) = x^{2} + 8, x \geq 0

x - 8, - x - 8

Solution steps

y = x^{2} + 8

Replace

x

with

y

x = y^{2} + 8

Solve for

y, x = y^{2} + 8

x - 8, - x - 8

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

f (x) = - 5 x^{3} - 2 x^{4}

f (x) = 4 x + 9

\frac{4 - x}{( x + 1 ) ( x ^{2} + 1 )}

f (x) = (x + 20)^{2} - 30

What is the inverse of f(x)=x^2+8,x>= 0 ?
The inverse of f(x)=x^2+8,x>= 0 is sqrt(x-8),-sqrt(x-8)