1. ## inequalities

Sketch on the same coordinate axes , the graphs for $y=\frac{1}{|x|}$ and $4y=3|x|-4$

Hence , find the set of values of x such that $3|x|-4>\frac{1}{|x|}$

I managed to sketch $y=\frac{1}{|x|}$ and $y=\frac{3}{4}|x|-1$

Is $\frac{3}{4}|x|-1>\frac{1}{|x|}$ the same as $3|x|-4>\frac{1}{|x|}$ ??

Zz i am confused . Thanks !

2. Originally Posted by thereddevils

Hence , find the set of values of x such that $3|x|-4>\frac{1}{|x|}$

I managed to sketch $y=\frac{1}{|x|}$ and $y=\frac{3}{4}|x|-1$

Is $\frac{3}{4}|x|-1>\frac{1}{|x|}$ the same as $3|x|-4>\frac{1}{|x|}$ ??

Zz i am confused . Thanks !
no, they can't be equal. It seems you multiplied by 4, but you need to do it to the entire inequality, not just one side, so the inequality becomes $3\left| x \right|-4>\frac{4}{\left| x \right|}.$

now i am confused, because dunno if you want to solve this inequality or the first one you wrote.

3. Originally Posted by Krizalid
no, they can't be equal. It seems you multiplied by 4, but you need to do it to the entire inequality, not just one side, so the inequality becomes $3\left| x \right|-4>\frac{4}{\left| x \right|}.$

now i am confused, because dunno if you want to solve this inequality or the first one you wrote.
Thanks so i will need to sketch $y=\frac{1}{|x|}$ and $4y=3|x|-4$ and find their intersection ?

4. no, you can solve it analitically, can you do that?

5. Originally Posted by Krizalid
no, you can sove it analitically, can you do that?
Yes . But i am not sure how to do it ?

6. Originally Posted by thereddevils
Thanks so i will need to sketch $y=\frac{1}{|x|}$ and $4y=3|x|-4$ and find their intersection ?
You need to find values of x so that the graph of y= 3|x|+ 4 is above the graph of y= 1/|x|.

7. Originally Posted by HallsofIvy
You need to find values of x so that the graph of y= 3|x|+ 4 is above the graph of y= 1/|x|.

why is it y=3|x|+4 , what happen to 4y=3|x|-4 ?? Thanks .

8. i gotta work now.

when i get back i'll tell you how to make it.

9. Originally Posted by thereddevils

Hence , find the set of values of x such that $3|x|-4>\frac{1}{|x|}$
is this the inequality you need to solve?

10. Originally Posted by Krizalid
is this the inequality you need to solve?
yes

11. okay, sorry for the delay.

here's the solution:

obviously it's $x\ne0$ and besides $|x|>0,$ so we rewrite the inequality as $3|x|^2-4|x|-1>0.$ This is not factorable as we want, so with some of algebra we get $3\left| x \right|^{2}-4\left| x \right|-1=\frac{9\left| x \right|^{2}-12\left| x \right|-3}{3}=\frac{\left( 3\left| x \right|-2 \right)^{2}-7}{3}>0,$ hence $
\left( 3\left| x \right|-2+\sqrt{7} \right)\left( 3\left| x \right|-2-\sqrt{7} \right)>0,$
and the first factor is positive, so we require that $3\left| x \right|-2-\sqrt{7}>0\implies x>\frac{2+\sqrt{7}}{3}\,\,\vee \,\,x<-\frac{2+\sqrt{7}}{3}.$