What is the inverse of f(x)=sqrt(x^2-9),x>= 3 ?

The inverse of f(x)=sqrt(x^2-9),x>= 3 is sqrt(x^2+9),-sqrt(x^2+9)

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inverse of f(x)=sqrt(x^2-9),x>= 3

inverse

f (x) = x^{2} - 9, x \geq 3

x^{2} + 9, - x^{2} + 9

Solution steps

y = x^{2} - 9

Replace

x

with

y

x = y^{2} - 9

Solve for

y, x = y^{2} - 9

x^{2} + 9, - x^{2} + 9

2^{1 + \frac{1}{2}} x^{\frac{3}{2}}

f (x) = 2 (3^{x}) + 1

f (x) = - 13 x - 9

3 x + 9

f (x) = 3 8 x

What is the inverse of f(x)=sqrt(x^2-9),x>= 3 ?
The inverse of f(x)=sqrt(x^2-9),x>= 3 is sqrt(x^2+9),-sqrt(x^2+9)