# Set of values of X

• May 7th 2011, 07:12 AM
Punch
Set 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 AM
topsquark
Quote:

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

And what did your graph tell you?

-Dan
• May 7th 2011, 10:50 PM
Punch
Quote:

Originally Posted by topsquark
And what did your graph tell you?

-Dan

From my understanding, the 2 graphs would intersect at 2 points, but i doubt this could help in anyway. Is there something I overlooked?
• May 8th 2011, 04:36 AM
topsquark
Quote:

Originally Posted by Punch
From my understanding, the 2 graphs would intersect at 2 points, but i doubt this could help in anyway. Is there something I overlooked?

Quote:

Originally Posted by Punch
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 $\displaystyle \frac{2(x+a)}{x-1}\leqslant |x+a|$

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 AM
Punch
Quote:

Originally Posted by topsquark
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

but i cant find out since i cant draw the graph of |x+a| without knowing the value of a