What is the inverse of f(x)=sqrt(x^2-1),x>= 1 ?

The inverse of f(x)=sqrt(x^2-1),x>= 1 is sqrt(x^2+1),-sqrt(x^2+1)

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inverse of f(x)=sqrt(x^2-1),x>= 1

inverse

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

x^{2} + 1, - x^{2} + 1

Solution steps

y = x^{2} - 1

Replace

x

with

y

x = y^{2} - 1

Solve for

y, x = y^{2} - 1

x^{2} + 1, - x^{2} + 1

y = \frac{1}{x - 3}

f (x) = 0.5 (1 - e^{2 x})

f (x) = lo g_{2} (x + 7)

y = 2 (3 x + 1) + 4

f (x) = ln (x^{59})

What is the inverse of f(x)=sqrt(x^2-1),x>= 1 ?
The inverse of f(x)=sqrt(x^2-1),x>= 1 is sqrt(x^2+1),-sqrt(x^2+1)