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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.

Fadzlin Rafi
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  • How far have you gotten in solving the problems? – GTX OC Oct 25 '13 at 16:01
  • I have no idea in question 1. But in question 2, I got 2 answers for part i) however the question stated find the value of k. And for part ii) I tried letting g(x1) =/= g(x2), not sure if its correct. – Fadzlin Rafi Oct 25 '13 at 16:16

1 Answers1

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(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

mert
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