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

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

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inverse of f(x)=x^2-16,x<infinity

inverse

f (x) = x^{2} - 16, x < \infty

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

- 3 x^{2} + 3 x^{2} + 6

f (x) = - 1 + x^{2}

\frac{x}{2 x - 3}

f (x) = 4 sin (x) + 3

f (x) = \frac{3 x + 8}{8 x - 7}

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