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I am really struggling on this exercise, and don't even know where to start.

(a) Use the fact that $|a|=\sqrt{a^2}$ to prove that, given $\epsilon>0$, there exists a polynomial $q(x)$ satisfying $||x|-q(x)|<\epsilon$ for all $x$ in $[-1,1]$.

(b) Generalize this conclusion to an arbitrary interval $[a,b]$.

I recognize that this is the conclusion of the Weierstrass approximation theorem. But I don't know how to use it.

Thank you so much in advance for your help!

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    Notice that the absolute value function is continuous. So it follows immediately from the theorem. – 2017-02-26
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    For an explicit construction, find a polynomial $p$ such that $|p(x)-\sqrt x |<\epsilon$ for $x\in [0,1].$ Then for $x\in [-1,1]$ you have $|p(x^2)-|x|\;|=|p(x^2)-\sqrt {x^2}|<\epsilon.$ – 2017-02-26
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    But if I simply say "the absolute value function is continuous, so the the WAT says ||x|−q(x)|<ϵ", is it enough? I mean, they are asking on [−1,1] first an then on any [a,b]... It can not be that simple right? – 2017-02-26
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    And if I use you reasoning with leading to |p(x2)−x2‾‾√|<ϵ... Am I suppose to find an explicit expression of p(x)? I think I am really confuse about what the problem is asking... – 2017-02-26
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    The exercise as written is just asking for the assurance of the existence of such polynomial. We do not need to know the explicit formula of such polynomial. That is, this is just a straightfoward implication of WAT. What statement of WAT you have in mind? – 2017-02-27
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    Well my textbook says "if f:[a,b]->R is continuous, then given ϵ>0, there exists a polynomial p(x) satisfying |f(x)−p(x)|<ϵ for all x in [a,b]." So, if I understand what you say, I am suit suppose to prove that the absolute value is continuous on [-1,1] and then on [a,b]? – 2017-02-27
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    You understood correctly. If the hypothesis of theorem is verified, then you can claim as true its implication. Hence, it suffices to prove that the absolute value function is continuous on any interval. Then you can conclude by WAT what the exercise is claiming. – 2017-02-27
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    Thank you!! To use the hint given by the exercise, I was thinking about proving that the function is continuous as a composition of 2 continuous functions: x^2 and sqrt(x). However, sqrt(x) is not defined on (-1,1), it is actually only defined on R+... So this can not prove the continuity of the absolute value neither on (-1,1) or on (a,b), right? – 2017-02-27
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    Actually this approach is perfectly fine. Now you should look at the image of $x^2$ and the domain of $\sqrt{x}$. I am sure you can complete the reasoning by yourself. – 2017-02-27

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