1. ## Inequality question

Hey guys, again i need help with the 2nd part of the question:

Find the range of values of x for which satisfy the inequality 1/(x-1) > 2/(x+2). Hence deduce the values of x for 1/(1-e^x) > 2/(1+2e^x).

Thanks!

2. Originally Posted by margaritas
Hey guys, again i need help with the 2nd part of the question:

Find the range of values of x for which satisfy the inequality 1/(x-1) > 2/(x+2). Hence deduce the values of x for 1/(1-e^x) > 2/(1+2e^x).

Thanks!
$\displaystyle \frac{1}{x - 1} > \frac{2}{x + 2}$

$\displaystyle \frac{1}{x - 1} - \frac{2}{x + 2} > 0$

$\displaystyle \frac{(x + 2) - 2(x + 1)}{(x - 1)(x + 2)} > 0$

$\displaystyle \frac{-x}{(x - 1)(x + 2)} > 0$

Now, for what values of x is this true? Critical values of x are where the numerator or denominator is 0, so let's split the real line into 4 segements and check:
$\displaystyle ( -\infty, -2): \frac{-x}{(x - 1)(x + 2)} > 0$ (Check!)

$\displaystyle ( -2, 0): \frac{-x}{(x - 1)(x + 2)} < 0$ (No!)

$\displaystyle ( 0, 1): \frac{-x}{(x - 1)(x + 2)} > 0$ (Check!)

$\displaystyle ( 1, \infty): \frac{-x}{(x - 1)(x + 2)} < 0$ (No!)

So the solution set is:
$\displaystyle (-\infty, -2) \cup (0, 1)$

Can you do the rest?

-Dan

3. Erh yes I got the answer for the first part but I still can't do the 2nd part.

4. Hi:

Using the identical procedure as did Dan, I get solution set (Ln(0.25), 0).

Regards,

Rich B.