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