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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.

  • 1
    $|x| = |-10| = 10$ is the maximum actually. – 2017-02-09
  • 0
    Oops, I forgot to ignore the negative sign on -10. Would I be still be technically correct since it just asks for a number that would satisfy the inequality? Ignoring fact that I mistook 9 as the largest. – 2017-02-09

1 Answers 1

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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).