If $-2 \leq x \leq \pi/2$, show that $|2x^3 - 4x^2 + 3x - \sin x|\leq 39$

can someone help me with this question? i'm having difficulty trying to incorporate triangle inequality with it.

  • 1
    $\begingroup$ What is the question, at the moment you only have "if $-2$"? $\endgroup$ – Thomas Russell Jan 17 '14 at 10:04
  • $\begingroup$ Sorry I don't know how to use symbols so I typed it $\endgroup$ – user120709 Jan 17 '14 at 10:05

Since $|x|\le 2, |\sin x|\le 1,$ $$\begin{align}|2x^3-4x^2+3x-\sin x|&\le 2|x|^3+4|x|^2+3|x|+|\sin x|\\&\le 2\cdot 2^3+4\cdot 2^2+3\cdot 2+1\\&=39.\end{align}$$

(Note that there is no $x$ such that the equality holds. This implies that $39$ is not the max.)

In general, $$|a+b|\le |a|+|b|,$$ $$|a-b|\le |a|+|b|.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.