Solution
= Solution
Every function with the stated two linear pieces can be written
$$
f(x)=a+bx+c\left|x-\frac12\right|
$$
for suitable real constants $a,b,c$. Since
$$
\left|x-\frac12\right|=\sqrt{\left(x-\frac12\right)^2},
$$
define
$$
Q_i(x)=a+bx+cP_i\!\left(\left(x-\frac12\right)^2\right).
$$
This is a polynomial. The argument of $P_i$ lies in $[0,1/4]\subseteq[0,1]$, so
$$
\boxed{\|Q_i-f\|_\infty
\leq |c|\sup_{u\in[0,1]}|P_i(u)-\sqrt u|
\longrightarrow0.}
$$