Why does $g(x)=(x^2−1)/3$ map $[-1,1]$ to $[-1/3, 0]$?

Question

This is a particular part of a fixed-point problem I don't understand. The problem says:

Let $g(x)=\dfrac{(x^2−1)}3$. Does $g$ have a fixed point on the interval $[−1,1]$?

So naturally we have to see what $g$ maps $[-1,1]$ to. My professor says it maps to $[-1/3, 0]$. My understanding is that, in general, if we have an interval $[a,b]$, then $g$ would map this to $[\min(g(a),g(b)), \max(g(a),g(b))]$. In which case, $g$ does not map $[-1,1]$ to $[-1/3, 0]$, but to $[g(-1),g(1)]=[0,0]$ instead.

If I am correct in what I say, then this means my professor has made a mistake and means to say the interval in question is $[0,1]$, which does indeed map to $[-1/3,0]$.

What is going on here?

Answer 1

Your understanding is wrong; in general, a function doesn't even necessarily map an interval to an interval, let alone the interval you mention. A continuous function does map intervals to intervals. But it is not true that the interval $[a,b]$ is then mapped to $[\min\{g(a),g(b)\},\max\{g(a),g(b)\}]$. Instead it is mapped to $$\left[\min_{x\in[a,b]}g(x),\max_{x\in[a,b]}g(x)\right].$$ That is, your should consider the values of $g$ on the entire interval, not just the endpoints.

As a side note to your original question; you don't need to see where $g$ maps the interval $[-1,1]$ to. You just need to ask yourself whether there exists some $x\in[-1,1]$ such that $g(x)=x$.

Answer 2

You make a mistake here, the plot of $g$, it decreases in $[-1,0]$ from 0 to $-\frac{1}{3}$, then increases.

Answer 3

That's immediate it is an even function, so the image is the same as the image of $[0,1]$, and on the latter interval, $g$ is monotonically increasing.