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Find a number M such that $\mid x^3 - x^2 +8x\mid \leq M$ for all $-2\leq x \leq 10$

I am not sure my answer is right or not..

I used triangle inequality.

$\mid x^3 - x^2 +8x\mid \leq \mid x^3 \mid + \mid x^2 -8x \mid \leq \mid x^3 \mid + \mid x^2 \mid + \mid 8x \mid $

$M = 10^3 + 10^2 + 8(10)$

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    Good job, that's a great approach. – 2017-02-10
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    Is this right way to solve it ? ..? – 2017-02-10
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    There are others, but this is one. If there is supposed to be a "right" answer only your teacher would be able to tell you what he/she is looking for. – 2017-02-10
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    Or, you could use Calculus to find the max and min values of $x^3-x^2+8x$ on the interval $[-2,10]$. – 2017-02-10

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Isn't the best result |$ 10^3 - 10^2 +8*10$|?see the image here