Proof with triangle inequality Find a number $Y$ such that $|x^5 - 3x^4 + 2x^3 - x| \leq Y$ for all $x \in [-10, 9]$.
My attempt:
By triangle inequality it follows that 
$|x^5 - 3x^4 + 2x^3 - x| \leq |x^5| + |3x^4| + |2x^3| + |x|$
$= |x|^5 + 3|x|^4 + 2|x^3| + |x|$
It is obvious that $|x|^5 + 3|x|^4 + 2|x^3| + |x|$ is largest when $|x|$ is largest. In the interval provided, $|x|$ is largest when x = -10 and so $|x| = 10$. One possibility for $Y$ is $Y = 10^5 + 3(10)^4 + 2(10)^3 + 10 = 132010$.
Is what I have here correct? If not where did I go wrong?
Edit: 9's were changed to 10's as according to comments.

Yes, that is correct.  Typically this method would offer quite a crude estimate due to ignoring cancellation that may occur.  However, in this case, because of the signs of the coefficients, all but the $-x$ term will combine to greatest magnitude when $x=-10$, so that the original absolute value is only $20$ smaller than the bound you found when $x=-10$.  So you couldn't do much better in this case (your $Y$ is about $0.015$% larger than the smallest possible bound).
