|x| ≤ 1 ⇒ |x^2 − x − 2| ≤ 3|x + 1|

im guessing you factor it and cancel out the x+1 then im left with this

|x-2| ≤ 3

What do i do from there?

Printable View

- Sep 8th 2010, 04:48 PMcalculuskid1Simple proving the inequalities
|x| ≤ 1 ⇒ |x^2 − x − 2| ≤ 3|x + 1|

im guessing you factor it and cancel out the x+1 then im left with this

|x-2| ≤ 3

What do i do from there? - Sep 8th 2010, 05:29 PMProve It
$\displaystyle |x - 2| = |x + (-2)|$.

Now

$\displaystyle |x + (-2)| \leq |x| + |(-2)|$ by the triangle inequality.

$\displaystyle |x + (-2)| \leq |x| + 2$.

And since $\displaystyle |x| \leq 1$

that means $\displaystyle |x| + 2 \leq 3$.

Therefore $\displaystyle |x - 2| \leq 3$

and thus

$\displaystyle |x| \leq 1 \implies |x^2 - x - 2| \leq 3|x + 1|$.