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

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

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inverse of f(x)=x^2-16,x>= 0

inverse

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

x + 16, - x + 16

Solution steps

y = x^{2} - 16

Replace

x

with

y

x = y^{2} - 16

Solve for

y, x = y^{2} - 16

x + 16, - x + 16

- 4 x + 1

f (x) = (5 - x^{3})^{\frac{1}{5}}

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

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

f (x) = \frac{4 x + 6}{5}

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