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I have some trouble to figure out the plot/graph of a set like $$M:=\{(x,y) \in R:\frac{1}{2}\leq x \land 1<y<\frac{1}{x}\}$$

On wolframalpha I saw the plot.

Is there a trick how to figure out such kind of graph?

Greetings.

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Graph all the points $(x,y)$ such that $1/2 \leq x$, then do the same for the points with $1<y$ and $y<1/x$, then consider the intersection of these three regions. –  01000100 Aug 18 '12 at 11:12
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I suppose there's no such trick. Simply, you have to make as much effort as you can to get the more comprehensive view in the structure of the set you are asked to plot.

In this case, I suggest you to start with the fixed range of variables, namely $x\ge \frac{1}{2}$ and $y\gt1$. You see that the plane was divided into the quarters. The one you are interested in is the up-right. Then you can move to plotting the more sophisticated part, i.e. $y\lt \frac{1}{x}$. In order to do that, sketch the $y=\frac{1}{x}$ graph and mark the underlying part of the plane. Once you've finished drawing all parts, you have to think what kind of conjugates were used to describe the set. Notice that in your example one has only "and" operators. Therefore, to obtain the result you have to take an intersection of all previously driven components.

Don't forget to exclude points which were probably marked by drawing boundaries, as it happens often when sketching sets defined with strict inequalities. You could do that by drawing all "edges" with dashed line first and then bolding ones which reflect weak inequalities. In this case, $y=1$ and $y=\frac{1}{x}$ should be excluded.

Edit: To obtain the graph of $y=\frac{1}{x}$, you can start with calculating coordinates of the points from the graph lying on the boundaries of the up-right quarter, namely two points of the form $(x,\frac{1}{x})$, where first has $x$ coordinate equal to $\frac{1}{2}$ and second $y$ one equal to $1$. It provides you with the points $(\frac{1}{2},2)$ and $(1,1)$. The remaining question is: What is the curve I am supposed to draw between the points? If you want the curve to be accurate, you have to be sure about such its features like monotonicity, concavity/convexity... All of them can be termined with tools known from calculus. Let me stop here as I'm not sure if it was exactly what you were asking about.

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