Inequalities « Mathematics Hothouse

Problem:

Graph the functions $f(x)=\frac{3}{x-1}$ and $\frac{2}{x+1}$ together to identify the values of x for which we can get $\frac{3}{x-1} < \frac{2}{x+1}$

Also, confirm your findings in the above graph algebraically/analytically.

Also, then find the points of intersection of these graphs using a TI graphing calculator and also algebraically/analytically.

Solution:

Just a suggestion, it helps to have TI graphing calculator, a bit pricey, but …it might help a lot…If you wish, you can grab one from the following:

http://www.amazon.in/TEXAS-GRAPHIC-CALCULATOR-NSPIRE-CX/dp/B00A49F98U/ref=sr_1_fkmr0_1?s=electronics&ie=UTF8&qid=1493101690&sr=1-1-fkmr0&keywords=TI+graphing+calculator+Non+CAS

Of course, there might be excellent freeware software packages that plot graphs, but I am strongly biased: I like the TI graphing calculator toy very much!! I go to the extent of claiming that it is better for a kid to have a nice graphing calculator like a TI than a smartphone! 🙂 🙂 🙂

Later on, I hope to show how a detailed study of graphs helps right from high school level. In fact, an immortal Russian mathematician, I. M. Gel’fand had written a book titled “Functions and Graphs” for high-school children. You can check if it is available in Amazon India or Flipkart or Infibeam, etc.

Oh, one more thing: I love this topic of inequalities because of several reasons. Just now, one of my students, Darpan Gajra asked me certain questions about inequalities when he tried to solve the above problem —- I think his questions were good, and I hope this little explanation that I gave him helps many of other students also.

Explanation:

Consider the given inequality: $\frac{3}{x-1} < \frac{2}{x+1}$ .

You might be tempted to solve this very fast in the following way:

Just cross-multiply: So, we get $3(x+1)<2(x-1)$ , which in turn means $3x+3<2x-2$ , that is, $x<-5$ . This is the answer by “fluke”!!!

In fact, the problem requires a detailed solution as follows:

Firstly, let us see what is wrong with the above fast solution:

Consider any inequality $a<b$ . Now, we know that only if $x>0$ , then $ax<bx$ . But, if $x<0$ , then $ax>bx$ . Now, in the present question, if you simply cross-multiply, you are not clearing assuming whether $(x-1)>0$ or $(x-1)<0$ ; $(x+1)>0$ or $(x+1)<0$ .

Also, one simple yet, I call it a golden rule is: even though it is not mentioned explicitly in the question, when you solve the question, immediately write the big restrictions: $x \neq 1$ , $x \neq -1$ as these values make the denominators of the two fractions zero. Develop this good habit.

So, now, coming to the solution:

$\frac{3}{x-1} < \frac{2}{x+1}$ , where $x \neq \pm 1$

$\frac{3}{x-1} - \frac{2}{x+1}<0$

$\frac{3x+3-2x+2}{x^{2}-1}<0$

$\frac{x+5}{x^{2}-1}<0$

Case I: $(x^{2}-1)>0$ : Multiplying both sides of the above fraction inequality by $(x^{2}-1)$ gives us:

$x+5<0$ , that is, $x<-5$ , but also $(x^{2}-1)>0$ is a restriction. So, it means that $x<-5$ AND $(x-1)(x+1)>0$ .

Subcase Ia: $(x+1)>0$ and $(x-1)>0$ , which gives, $x>-1$ and $x>1$ , which together imply that $x>1$ , but we need $x<-5$ also as a restriction/assumption. So, in this subcase Ia, solution set is empty.

Subcase Ib: $x+1<0$ and $x-1<0$ , which in turn imply, $x<-1$ and $x<1$ , that is, $x<-1$ , but we need $x<-5$ also. Hence, in subcase Ib: $x<-5$ and $x<-1$ , we get $x<-5$ as the solution set.

Case II: $(x^{2}-1)<0$ : Multiplying both sides of $\frac{x+5}{x^{2}-1}<0$ by $(x^{2}-1)$ , we get:

$(x+5)>0$ , hence, $x>-5$ . But, we also need the restriction/assumption: $(x^{2}-1)<0$ , which implies that $(x+1)(x-1)<0$ .

Subcase IIa: $(x+1)>0$ and $(x-1)<0$ , that is, $-1<x<1$

So, in subcase IIa, we have $(x>-5) \bigcap (-1<x<1)$ , that is, $(-1<x<1)$ .

Subcase IIb: $(x+1)<0$ , and $(x-1)>0$ , that is, $x<-1$ and $x>1$ . But, this itself is an empty set. Hence, in subcase IIb, the solution set is empty.

Now, the final solution is $\{ case I\}$ OR $\{ case II\}$ , that is, $\{ case I\} \bigcup \{ case II\}$ , that is, $\{ x<-5\}\bigcup {-1<x<1}$