MHB Rational Inequalities: Solve & Understand | Math

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The discussion focuses on solving the inequality \( \frac{1}{x} > 2 \) by analyzing different cases based on the sign of \( x \). It emphasizes the importance of considering the direction of the inequality when multiplying by a variable, particularly when \( x \) is negative. The solution is derived by determining intervals: for \( x > 0 \), the valid range is \( 0 < x < \frac{1}{2} \), while no values for \( x < 0 \) satisfy the inequality. The method involves checking values within the identified intervals to confirm which satisfy the original inequality. Ultimately, the conclusion is that the solution set is \( 0 < x < \frac{1}{2} \).
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Uhh ... fill in the blanks? Have you worked on any of these?
 
skeeter said:
Uhh ... fill in the blanks? Have you worked on any of these?
Idk how the solution works
 
for $\dfrac{1}{x} > 2$

step 1. $\dfrac{1}{x} - \dfrac{2}{1} > 0$

combine the two fractions by using a common denominator
 
Personally, I wouldn't do it that say.
From $\frac{1}{x}> 2%$, multiply on both sides by x.
But you have to be careful with that! Unlike with an equation, multiplying on both sides by negative number reverses the ">" sign. So do two cases:

1) If x> 0 then $1>2x$. Divide on both sides by the positive number 2: $\frac{1}{2}> x$..
Since we are requiring that x be positve, we have $0< x< \frac{1}{2}$

2) If x< 0 then $1< 2x$. Divide on both sides by the positive number 2: $\frac{1}{2}< x$, But since we are requiring that x be negative, that is not possible.

The solution is $0<x \frac{1}{2}$.

It is also true that, for continuous functions, g and f, to change from f(x)<g(x) to f(x)> g(x), we have to go through f(x)= g(x) or a poinr where either f or g is undefined.

So start by solving the equation $\frac{1}{x}= 2$. That is the same as $1= 2x$, or $x=\frac{1}{2}$. I is also true that $\frac{1}{x}$ is undefined for x= 0. That divides the real numbers into three intervals, x< 0, 0< x< 1/2, and x> 1/2. We need only check one value of x in each interval. For x< 0 take x=-1. Then 1/x= -1 which is NOT larger than 2 so no x less than 0 satisfies 1/x> 2. For 0< x< 1/2 we can take x=1/4. Then 1/x= 4 which is greater than 2. Every number betwen 0 and 1/2 satisfies the inequalty. Finally take x= 1. Then 1/x=1 which is not larger than 2. No x larger than 1/2 satisfies the inequality.
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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