Finding the number of solutions to an equation given the derivative 
Today my teacher gave us this question in class to think about at home. I've been through my textbook trying to find a way to answer this but I have no idea how to find the number of solutions to $g(x)=0$ given that I only have the graph of its derivative, $g'$.
 A: Actually, there can possibly be up to $2$ solutions for $g(x) = 0$, as the graph of $g(x)$ is basically of a concave up parabola with a vertex minimum at $x = 0$. Thus, if the critical point of $g'(0) = 0$ occurs where $g(0) \lt 0$, then as $g'(x) \lt 0$ for $x \lt 0$, this means $g(x)$ is decreasing for $x \lt 0$ and, thus, it could cross from above to below the $x$ axis somewhere in $(a,0)$. Similarly, as $g'(x) \gt 0$ for $x \gt 0$, this means $g(x)$ is increasing for $x \gt 0$ and, as such, it could cross from below to above the $x$ axis somewhere in $(0,b)$.
However, if $g(0) = 0$, then there will be only $1$ zero, and if $g(0) \gt 0$, there will only be no zeros.
As there are no other critical points, with $g(x)$ strictly monotonically decreasing for $x \in [a,0)$, and strictly monotonically increasing for $x \in (0,b]$, there can be no more than $2$ points where $g(x) = 0$ for $x \in [a,b]$.
A: Looks like $0,1$ or $2$ zeros for $g$.
Consider for instance $g(x)=x^2$.  It has $1$ zero.
Or $g(x)=x^2+1$. It has none.
Or $g(x)=x^2-1$. It has two.
We could also subtract a large enough constant $c$ so that $g(x)=x^2-c$ has no zeroes on $[a,b]$.
$g$ will resemble one of these on $[a,b]$. 
