If $f(x)$ is differentiable in $\mathbb{R}$ & $a,b \in \mathbb{R}, a\neq b$ such that $f'(x)=(x-a)(x-b)$ then $f$ has exactly one local min and max? Prove or contradict: 

If $f(x)$ is differentiable in $\mathbb{R}$ and exist $a,b \in \mathbb{R}, a\neq b$ such that $f'(x)=(x-a)(x-b)$ then $f$ has exactly one local minimum and 1 local maximum

I know that $f'$ has only two roots which are $a$ and $b$ as the possible locations for min and max points, but how do I show these are necessarily or are not necessarily min and max points?
 A: Let $a<b$ without loss of generality. For $x<a$ we have $f'(x)>0$, for $x=a$ we have $f(a)=0$ and for $a<x<b$ we get $f'(x)<0$. This can be interpreted as $f$ is increasing up to $x=a$ and then decreases while $x$ approaches $b$ from the left. So $f(a)$ is a maximum. Similarly $f'(b)=0$ and $f'(x)>0$ for all $x>b$ which implies that $f$ once it reaches $f(b)$ starts increasing. So $f(b)$ is a minimum. These two points $a,b$ are the only local extrema since $f'(x)$ vanishes exactly there by its very definition.
A: \begin{align} f''(x) = \left( x - b \right) + \left( x - a  \right) \end{align}
\begin{align} f''(a) = \left( a-b \right) \end{align}
\begin{align} f''(b) = \left( b -a \right) \end{align}
When \begin{align} a \gt b \end{align}
Then \begin{align} f''(a) \gt 0 \end{align}
And \begin{align} f''(b) \lt 0 \end{align}
When \begin{align} b \gt a \end{align}
Then \begin{align} f''(a) \lt 0 \end{align}
And \begin{align} f''(b) \gt 0 \end{align}
As you can see in both cases we have a local Maxima and minima
A: The stationary point qualification is done with 
$$
f''(x) = (x-a)+(x-b) = 2x-(a+b)
$$
hence we have
$$
f''(a) = a-b\\
f''(b) = -(a-b)
$$
so one of then is a relative minimum and the other a relative maximum deppending on $a > b$ or $b > a$
