Check you zero's again: $x_1 = -1$ is a zero, as is $x_2 = -5/3$.
Try evaluating $f(x)$ at $x = -3/2$, i.e., $x = -2$ and compare with find $f(-5/3)$
Likewise for $f(-1)$. Choose smaller intervals around each critical point. Try, say, evaluating $f(x)$ at $x = -3/2$ and $x = -1/2$, to compare with $f(-1)$.
Since you have two critical points with only $2/3$ of a unit separating them, you need smaller intervals to determine the behavior of the function near those point.
You can also use the sign of the derivative to determine on which interval(s) a function is increasing, and when it is decreasing. When $f'(x) > 0 \implies f(x)$ is increasing, when $f'(x) \lt 0 \implies f(x) $ is decreasing. But again, you'll want to evaluate $f'(x)$ for $x$ very near the critical points $x_1, x_2$
If $x$ is rational, so $f(x)=x$, are there values $y$, as close as you like to $x$, with $f(y)>f(x)$? Are there values $z$, as close as you like to $x$, with $f(z)<f(x)$?.
Then ask the same questions for irrational $x$.
Best Answer
What about
$$f(x)=\begin{cases} x^2\sin\left(\frac{1}{x}\right) & x \neq 0\\ 0 & x=0 \end{cases}$$
at $x=0$?