Find a number which satisfies the equation $|x^6-3x^3+5x| \leq Q$ whenever $|x|$ $\leq$ $2$? NOTE (EDIT): I have noticed that no one noticed that a TRIANGLE INEQUALITY is being used. Hence, part of the confusion in my comments of the solution. 
TRIANGLE INEQUALITY: 
$|x+y| \leq |x|+|y|$, important part to note that there is $\leq$ symbol.  
So there this question in which a number has to be found such that it satisfys the following equation (whenever $|x|$ $\leq$ $2$): 
Also: this is strictly a proofs question and no calculus is allowed. Thanks.
$|x^6-3x^3+5x| \leq Q$ whenever $|x|$ $\leq$ $2$
$(1)$ $\leq |x^6+(-3x^3)+5x|$
$(2)$ $\leq |x^6|+|-3x^3|+|5x|$
$(3)$ $\leq |x|^6+3*|x|^3+5*|x|$ 
$(4)$ $\leq 2^6 + 3*2^3 + 5*2 = 98$
The question is: what is the problem asking for? Does it say to find a number in which it satisfies an expression between the interval of $x$ in $[-2, 2]$? Why do you substitute $|x|$ with $2$ specifically and not any other numbers? And, apparently, this solution can be done in another way to end up with another answer... because any number higher than 98 is also the answer? Thanks. 

Yes we have that for $|x|\le 2$
$$0-24-10\le x^6-3x^3+5x\le  64+24+10 \implies |x^6-3x^3+5x|\le 98$$
Note that the problem is not asking for the "minimum value $Q$ such that..." but simply for any value $Q$ which satisfy the inequality. Therefore the solution $Q=98$, even if it is not the minimal possible answer, is correct as any other $Q\ge 98$.
