Is the square of a strictly increasing (or decreasing) function strict quasi-convex?

Question

I was doing research on nonlinear channel estimation, where I need to consider the uniqueness of the optimal solution of the least-square problem under nonlinearity.

I am now thinking about one dimension case. Specifically, I need to solve the following optimization problem:

$\hat{x}=\arg\min_x \frac12(y-f(wx))^2$,

where $y$ and $w$ are some constant real numbers and $w\neq0$. When $f$ is the identity function, the problem has a unique solution at $\hat{x}=y/w$. Now suppose

1. $f$ is some strictly increasing function,
2. $f(x)$ is an odd function w.r.t. $x$,

does it still has a unique optimal solution (if it exists)?

I think if $f$ is strictly increasing function, the objective function seems to be strict quasiconvex, and thus any local optimal is unique global optimal. But I would like to know how to prove it is strict-quasi convex (if it is). Thank you.