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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.

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    $\begingroup$ 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
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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, $$|a+b|\le |a|+|b|,$$ $$|a-b|\le |a|+|b|.$$

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