Range of $f(z)=|1+z|+|1-z+z^2|$ when $ |z|=1$ 
Find the Range of $f(z)=|1+z|+|1-z+z^2|$ when $z$ is a complex $ |z|=1$

I am able to get some weaker bounds using triangle inequality.
$f(z)< 1+|z|+1+|z|+|z^2|=5$
also $f(z)>|1+z+1-z+z^2|=|z^2+2|>||z^2|-2|=1$,but
these are too weak!.as i have checked  with WA
Is there an elgant solution using minimum calculus?. Substituiting $x+iy$ makes things very complex indeed.
Answer:
$[\sqrt{\frac{7}{2}},3\sqrt{\frac{7}{6}}]$

Since $|z|=1$ we substitute $z=e^{i\theta}$, to get:
$$
f(\theta)=|1+e^{i\theta}|+|1-e^{i\theta}+e^{2i\theta}|\\
= \sqrt{ (1 + \cos \theta)^2 + \sin^2\theta} + \sqrt{(1 - \cos\theta+ \cos2\theta)^2 + (-\sin\theta + \sin2\theta )^2)}\\
=\sqrt{|2+2\cos\theta |}+|(1-2\cos\theta)|
$$
We can denote $u=\cos\theta$ and investigate $f(u)=\sqrt{|2+2u |}+|(1-2u)|$ for $u$ between $-1$ and $1$ by simple calculus. We get
$\max f=\frac{13}{4}$ for $u=-\frac{7}{8}$, and $\min f=\sqrt3$ for $u=\frac{1}{2}$. This does not agree with the answer given but I could not find any error.
