Hello i got problem with nr 62. I got some progress but i dont understand

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- Feb 13th 2013, 04:46 AMPetrusEpsilon
Hello i got problem with nr 62. I got some progress but i dont understand

- Feb 13th 2013, 06:50 AMemakarovRe: Epsilon
Maybe you can start by determining N from the graph instead of algebraically. Post your answers here.

- Feb 13th 2013, 07:12 AMPetrusRe: Epsilon
Hi im currently on with phone but it looks around when n=5 kinda hard to se the graf

- Feb 13th 2013, 07:20 AMemakarovRe: Epsilon
Yes, for ε = 0.5, N = 5 and even N = 3 works.

So, you need to determine for which x > 0 it is the case that $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}>2-\varepsilon=2-0.5=1.5$. Multiply both sides by (two times) the denominator, take square of both sides and solve (approximately) the resulting quadratic equation. - Feb 13th 2013, 09:42 AMPetrusRe: Epsilon
- Feb 13th 2013, 09:55 AMemakarovRe: Epsilon
- Feb 13th 2013, 10:04 AMPetrusRe: Epsilon
i will try again, try se if i can se what i did wrong

- Feb 13th 2013, 10:27 AMPetrusRe: Epsilon
Here is one progress i made.. Im kinda confused right now[ATTACH=CONFIG]26998[/ATTACHAttachment 26999

- Feb 13th 2013, 10:38 AMemakarovRe: Epsilon
You start with the quadratic inequality, which skips a lot of steps. I don't agree with the inequality, but I can't point out the step with an error because these steps are skipped. I wrote a couple of intermediate results in post #6. I also recommended multiplying both sides by 2 (or 4) to get rid of fractions like 1.5.

Here is a sequence of steps I recommend. Start with $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}>1.5$.

1. Multiply both sides by (x + 1). (The direction of the inequality does not change because we are looking for x > -1, where x + 1 > 0.)

2. Multiply both sides by 2 to get rid of 1.5.

3. Take square of both sides. (This could lead to appearance of spurious solutions, but we ignore this for now.)

4. Move everything to the left-hand side and add like terms.

5. Solve the quadratic equation obtained when > is replaced with =. Let's denote the solutions by $\displaystyle x_1$ and $\displaystyle x_2$ where $\displaystyle x_1 < 0$ and $\displaystyle x_2 > 0$.

6. Since the leading coefficient of the quadratic polynomial f(x) is positive and the inequality has the form f(x) > 0, the solutions to the inequality are $\displaystyle x < x_1$ and $\displaystyle x > x_2$. We are interested in $\displaystyle x_2$.

Edit: Sorry, I missed that there are two attached images and not one. I'm looking at the second one... - Feb 13th 2013, 10:47 AMPetrusRe: Epsilon
idk im following my book how it tell me :P that is when f(x)>0 and now ima solve for f(x)<0

- Feb 13th 2013, 10:55 AMemakarovRe: Epsilon
OK, I see that you started from $\displaystyle \left|\frac{\sqrt{4x^2+1}}{x+1}-2\right|<0.5$, which you converted into $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}-2<0.5$. In fact, if we denote $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}$ with z, $\displaystyle |z-2|<0.5$ is equivalent to $\displaystyle -0.5<z-2<0.5$. Adding 2 to all sides, we get $\displaystyle 1.5<z<2.5$. So you are right that $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}<2.5$ is a part of the original inequality. However, if you saw the graph from post #2, you must have seen that as x tends to infinity, the graph approaches 2 from

*below*. This means that $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}<2$ for positive x, so the inequality $\displaystyle \frac{\sqrt{4x^2+1}}{x+1}<2.5$ is automatically true for positive x. What we are interested in is the other part: $\displaystyle 1.5<\frac{\sqrt{4x^2+1}}{x+1}$. It becomes true only starting from some N > 0. I assumed you saw all this after post #2 and therefore I recommended solving

$\displaystyle \frac{\sqrt{4x^2+1}}{x+1}>2-\varepsilon=2-0.5=1.5$ (*)

in post #4. Later, in post #6, I wrote a couple of inequalities you obtain while solving (*). You ignored all this and started solving a different inequality. - Feb 13th 2013, 11:02 AMPetrusRe: Epsilon
- Feb 13th 2013, 11:05 AMemakarovRe: Epsilon
I also recommend clicking the "Reply With Quote" button under my posts to see how formulas are typed using LaTeX. Using LaTeX is not difficult, and you won't need to attach pictures. Also check out the LaTeX Help subforum on this site.

- Feb 13th 2013, 11:07 AMPetrusRe: Epsilon
Defination 7 says me what im doing idk, but the problem says i shall use defination 7. Attachment 27001

- Feb 13th 2013, 11:09 AMPetrusRe: Epsilon