Functions

$f(x) = x^2 - 2x$ with domain x ≥ 1

Find:

a) the inverse
b) state the range of the inverse

b) The range of the inverse is the domain of f(x) which is x ≥ 1

Need assistance with part a)

Completing the square gives the inverse as $f^{-1}(x) = 1 \pm \sqrt{x + 1}$

How do I know to take the positive or negative square root as the inverse?

The function g(x) = 1 - sqrt(1+x) (with domain x >= -1) gets very negative as x gets large, so its range certainly can't be x >= 1...

I assume you mean the function g(x) with range x ≥ -1 and not domain x ≥ -1?

$f(x) = x^2 - 2x$ with domain x ≥ 1

Find:

a) the inverse
b) state the range of the inverse

Need assistance with part a)

Completing the square gives the inverse as $f^{-1}(x) = 1 \pm \sqrt{x + 1}$

How do I know to take the positive or negative square root as the inverse?

b) The range of the inverse is the domain of f(x) which is x ≥ 1

No he means domain. He's asking you to think about how that function behaves when x takes on large values.

We don't know the domain of g(x) - in this case the inverse. We can figure it out by figuring out the range of f(x) but that would be extra work.

Let's say I plugged in random numbers, x = 35 into $f^{-1}(35) = 1 - \sqrt{35 + 1} = -5$. That is a negative image/range value, so clearly $f^{-1}(x) = 1 - \sqrt{x + 1}$ cannot be the inverse as the range found by just looking at f(x) would be invalid which cannot be the case.

What I'm curious about now my first statement in this post - the domain of g(x). Did he figure it out? Was it even necessary? Can't you just plug in random large values?