In the question below, I would have to solve for an upper estimate using Weierstrass Approximation Theorem, however I am not familiar with the theorem, how do I go about solving it?
For any $\epsilon\in [0,1]$, find an upper estimate on the integer $n$ such that there exists an approximation of $f(x)=|x|$ on $[-1,1]$ by a polynomial $P(x)$ of degree $n$ such that $$\sup_{x\in[-1,1]}|P(x)-|x||\leq \epsilon.$$
One path for proving the Weierstrass theorem, that comes with a specific estimate, is to use the Bernstein polynomials $$B_n(f)(x) = 2^{-n}\sum_{k=0}^n f\left(\frac {2k-n}n\right)\cdot \binom{n}{k}(1+x)^k(1-x)^{n-k}$$ For uniformly continuous $f$ on $[-1,1]$, these converge uniformly to $f$ as $n\to\infty$. It's a probabilistic estimate; given a uniform continuity estimate $|f(x)-f(y)|\le g(t)$ whenever $|x-y|\le t$, we estimate \begin{align*}\left|B_n(f)(x)-f(x)\right| &= 2^{-n}\left|\sum_{k=0}^n \left(f\left(\frac {2k-n}n\right)-f(x)\right)\cdot \binom{n}{k}(1+x)^k(1-x)^{n-k}\right|\\ &=\sum_{k=0}^n\left|f\left(\frac {2k-n}n\right)-f(x)\right|\cdot 2^{-n}\binom{n}{k}(1+x)^k(1-x)^{n-k}\\ &\le \sum_{|(2k-n)/n-x|\le t}g(t)\cdot (*) + \sum_{|(2k-n)/n-x|> t}g(2)\cdot (*)\\ &\le g(t)\cdot P(X-x\le t) + g(2)\cdot P(X-x > t)\\ \left|B_n(f)(x)-f(x)\right| &\le g(t)\cdot 1 + \frac1{nt^2}g(2)\end{align*} In this, the random variable $X$ is the scaled binomial distribution with probability mass function $2^{-n}\binom{n}{k}(1+x)^k(1-x)^{n-k}$. That expression is also abbreviated with $(*)$ in the third line, for typographic reasons. This $X$ has mean $x$ and variance $\frac{(1+x)(1-x)}{4n}\le \frac1n$. For the final inequality, we estimate that the first probability is $\le 1$, and that the second probability is $\le \frac1{nt^2}$ by Chebyshev's inequality.
For the function $f(x)=|x|$ we're dealing with, we can take $g(t)=t$. The error estimate we get is then $$|B_n(f)(x)-f(x)|\le \inf_t\left(t+\frac{2}{nt^2}\right) = \left(\frac4n\right)^{\frac13} + \frac{2}{n\left(\frac4n\right)^{\frac23}} = \frac{3}{\sqrt[3]{2n}}$$ To make that less than $\epsilon$, we take $n>\frac12\cdot\left(\frac{3}{\epsilon}\right)^3$.
This is a theoretical estimate, that gives away more than it has to. It's also not the only way to find approximating polynomials. So, then, some follow-up questions: