How do I prove inequalities and one-to-one function? Can anyone please help me with these questions?
1.Given
x + 1 < 0
Prove that:
i) $2x - 1 < 0 $
ii) ${2x-1\over x+1} > 2$
2.For $g(x) = {kx + 8\over 4x - 5}$
i) Find k if gg(x) = x
Is it fine if I just let any value of x for this question?
ii) Find the value of k so that g(x) is not a one-to-one function.
Thank you very much.

(1 i) $x+1<0\Rightarrow 2\left( x+1 \right)<2.0\Rightarrow 2x+2<0\Rightarrow 2x+2-3<-3\Rightarrow 2x-1<-3<0$
so $2x-1<0.$
(1 ii)$$\frac{2x-1}{x+1}=\frac{2\left( x+1 \right)-3}{x+1}=2-\frac{3}{x+1}$$
because $x+1<0$ given we can write $$-\frac{3}{x+1}>0\Rightarrow 2-\frac{3}{x+1}>2 \Rightarrow\frac{2x-1}{x+1}>2$$
(2 i) we know if $g\left( x \right)=\frac{ax+b}{cx+d}$ then ${{g}^{-1}}\left( x \right)=\frac{-dx+b}{cx-a}$.
$g\left( g\left( x \right) \right)=x\Rightarrow g\left( x \right)={{g}^{-1}}\left( x \right)$ this means $g(x)=\frac{kx+8}{4x-5}=\frac{5x+8}{4x-k}={{g}^{-1}}\left( x \right)$ so $k=5.$ 
(2 ii) similar way
