Sketch the graph of $\displaystyle y=\frac{(2(x+a))}{(x-1)}$, where $\displaystyle a>2$. Hence find the set of values of x such that $\displaystyle \frac{2(x+a)}{x-1}\leqslant |x+a|$

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- May 7th 2011, 07:12 AMPunchSet of values of X
Sketch the graph of $\displaystyle y=\frac{(2(x+a))}{(x-1)}$, where $\displaystyle a>2$. Hence find the set of values of x such that $\displaystyle \frac{2(x+a)}{x-1}\leqslant |x+a|$

- May 7th 2011, 09:46 AMtopsquark
- May 7th 2011, 10:50 PMPunch
- May 8th 2011, 04:36 AMtopsquark
They intersect twice. Good. Now look at the inequality again. You are looking for all points on the 2(x + a)/(x - 1) curve that are

*lower*on the graph (or equal to) the graph of |x + a|. So what parts of the function 2(x + a)/(x - 1) are below the lines?

-Dan - May 8th 2011, 04:50 AMPunch