Let $f$ be a real valued continuous function on $[-1,1]$ such that $f(x)=f(-x)$ .Show that for every $\epsilon>0$ there is a polynomial $p(x)$ over $\Bbb Q$ such that $|f(x)-p(x^2)|<\epsilon$
Since $f(x)$ is a continuous function by Weierstrass Approximation there exists a polynomial $p(x)$ such that $|f(x)-p(x)|<\epsilon$.
Also since $\Bbb Q$ is dense in $\Bbb R$ so we can take $p(x)$ over $\Bbb Q$ and still have $|f(x)-p(x)|<\epsilon$.
Now $f(x)=f(-x), |f(-x)-p(-x)|<\epsilon\implies |f(x)-p(-x)|<\epsilon$
Thus we have $|f(x)-p(x)|<\epsilon$,$|f(x)-p(-x)|<\epsilon,x\in [-1,1]$
How to combine these two to get $|f(x)-p(x^2)|<\epsilon$? Please help.