Hi syndicate, thanks for your explanation. Yes, the way you did it is straightforward and more easier but can you also explain it the way they did it in the solution book please? I know you kinda touched on it a little but I'm still failing to understand the way they did it in the book. Thanks! 
So, the book has basically solved for x using the ratios from the triangles on the top.
Case 1: BC, in this case is the hypotenuse, and AB is the adjacent side from the angle.
AB has a ratio of \( \sqrt{3} \), where as BC has a ratio of 2. So BC: AB = \( 2: \sqrt{3} \)
Now the book has substituted the actual value of BC and AB into the ratios. BC = 20, and AB = x.
\( \therefore 20:x \space = \space 2:\sqrt3 \)
The book didn't show this trick, and converted these ratios into fractions (rational numbers).
y was calculated using this same method.
I had 2 answers when I tried to find f inverse which was -(rootx+1 +1) or positive of that
I just graphed xsquared +2x then to find my ranges and domain
Yes, when you find the inverse of g(x), you do get two answer, however, only one of them is valid.
^2 \\ \therefore y = -1 \space \pm\sqrt{x+1} )
The question states that the domain of g(x) is \( [-1, \infty) \), which means the range of \( g^{-1} \) is also \( [-1, \infty) \). Whereas \( y = -1 \space \pm\sqrt{x+1} \) has a range of \( ( - \infty, \infty) \) (it goes from -1 to -infinity and -1 to infinity). Since this function is already restricted, the correct answer is only \( \sqrt{x+1} -1 \), as it has a range of \( [- 1, \infty) \), where as \(- \sqrt{x+1} -1 \) would be considered an extraneous solution, as it has a range of \( (- \infty, -1] \) .
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