# Another beginners calculus problem

• Oct 28th 2010, 05:41 AM
jayshizwiz
Another beginners calculus problem
Find an $\displaystyle M>0$ so that every $\displaystyle x>M$ fulfills $\displaystyle \frac{6x^2-7\sin x}{3x+1} > 1000$

I've been brainstorming ideas, but it's a bit frustrating since I can't simplify the equation like I usually do.

Here are my thoughts and questions:

1) I'm thinking along the lines of proving something like $\displaystyle \frac{6x^2-7\sin x}{3x+1} > \frac{9x^2-1}{3x+1} > 1000$ because then I can simplify to $\displaystyle 3x-1>1000$. Is there any way I can make that happen?

2) Can $\displaystyle 6x^2-7\sin x$ ever be $\displaystyle < 0$. Because if not, we know that $\displaystyle 3x+1$ must also be$\displaystyle > 0$ since the function must be $\displaystyle > 1000$.

3) Or can I just say M = 1000000000000, and since sinx is always between 1 and -1, it doesn't effect us, and 6x^2 is always bigger than 3x+1. so a x>M is definitely bigger than 1000.

Thanks!
• Oct 28th 2010, 06:04 AM
Ackbeet
You'd like a sequence of inequalities that look like this:

$\displaystyle \dfrac{6x^2-7\sin(x)}{3x+1} >...>...>1000.$

So, if you can make the expression smaller but still with a size that you can control, you're good.

Since you're subtracting the sin function, and you want your function to get smaller, make the $\displaystyle 7\sin(x)$ term as big as possible:

$\displaystyle \dfrac{6x^2-7\sin(x)}{3x+1}\ge \dfrac{6x^2-7}{3x+1}.$

Now work with this, and make it bigger than the desired amount.
• Oct 28th 2010, 06:52 AM
jayshizwiz
Quote:

Originally Posted by Ackbeet
You'd like a sequence of inequalities that look like this:

$\displaystyle \dfrac{6x^2-7\sin(x)}{3x+1} >...>...>1000.$

So, if you can make the expression smaller but still with a size that you can control, you're good.

Since you're subtracting the sin function, and you want your function to get smaller, make the $\displaystyle 7\sin(x)$ term as big as possible:

$\displaystyle \dfrac{6x^2-7\sin(x)}{3x+1}\ge \dfrac{6x^2-7}{3x+1}.$

Now work with this, and make it bigger than the desired amount.

Thanks but I'm still stuck for some reason. Can you give me the next step?

I tried making $\displaystyle \dfrac{6x^2-7}{3x+1} > 1005 > 1000$ but I got stuck. I tried taking away a 1 and a 7 but I'm not so sure what to do.
• Oct 28th 2010, 06:57 AM
jayshizwiz
Hm...maybe I should simply solve for $\displaystyle \dfrac{6x^2-7}{3x+1} > 1000$

and I should get x>(something). Then I can just make M=(something) so I know x is always >M
• Oct 28th 2010, 07:05 AM
Ackbeet
I think that procedure should work for you.