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I was using the method of intervals to solve this Question. I seemed to miss $0$ on the number line while using the method of intervals. Why is $0$ included here in the method of intervals?

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$$\frac{2}{x}-3<0$$ or $$\frac{2-3x}{x}<0,$$ which gives the answer: $x>\frac{2}{3}$ or $x<0$.

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  • $\begingroup$ That's a really nice way to see it, by making a common denominator that forces us to look at the point of discontinuity. +1 $\endgroup$ – G Tony Jacobs May 9 '17 at 13:53
  • $\begingroup$ Could you make me understand why the answer is not x €(2/3, infinity) ? $\endgroup$ – user33699 May 9 '17 at 13:56
  • $\begingroup$ @user33699 We need to find all values of $x$, for which the inequality is true. For $x<0$ your inequality is also true and with $x>\frac{2}{3}$ it gives all solutions. $\endgroup$ – Michael Rozenberg May 9 '17 at 13:58
  • $\begingroup$ Can someone just tell me why 0 is included in the method of intervals ? $\endgroup$ – user33699 May 9 '17 at 14:24
  • $\begingroup$ @user33699 The other answer tells you that. The terminology is not quite right, what they probably mean is that the function is not defined at that point, so it can switch sign as it moves from left to right and still remain continuous. $\endgroup$ – Matt Samuel May 9 '17 at 14:36
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Because $\frac2x$ is not continuous at $x=0$. Note that just to the left of $0$, $\frac2x$ is large in magnitude but negative; while just to the right of $0$, it is large and positive.

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For an inequality $f(x) <0$ or $f(x)>0$, the method of intervals finds all the points where the graph of $f(x)$ crosses the $x$-axis (some people call these "split points".) Once you find them, they cut the $x$-axis into intervals. Each interval is either part of the solution set or not.

The big point is this: If one interval had values of $x$ that made $f(x) >0$ and other values of $x$ that made $f(x)<0$ then somehow the graph would have to cross the $x$-axis in that interval. But you know that it doesn't, because you have all the split points.

Finally, there are two ways for a graph to get to the other side of the $x$-axis. One is to intersect it (at such points $f(x) =0.$ The other way is to jump over it, (at such points $f(x)$ is discontinuous. So the collection of split points is the set of all points where either $f(x)=0$ or $f(x)$ is discontinuous.

For your problem we have $$\frac{2}{x}-3 <0.$$

So we solve $$\frac{2}{x}-3 = 0$$

and get $x=2/3$. Then we look for discontinuities and find $x=0$ is a problem.

The set of split points is $\{0, 2/3\}$ which cut the real line into the three intervals $(-\infty, 0)$, $(0,2/3)$, and $(2/3,\infty).$ I think you know the rest of the story.

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First suppose $x>0$. Then $$ \frac{2}{x}<3\iff3x>2\iff x>2/3. $$ If $x>0$ and $x>2/3$, then $x>2/3$. Now suppose that $x<0$. Now when we multiply by $x$ the inequality sign flips. Hence $$ \frac{2}{x}<3\iff3x<2\iff x<2/3. $$ If $x<0$ and $x<2/3$, then $x<0$. Thus the solution set is $(-\infty, 0)\cup(2/3,\infty)$.

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