$f(x)=(x-2)(x-7)(x-9)(x-12)$ how many roots is there to $f'(x)=0$ and $f''(x)=0$? without taking derivative.

Question

if $f(x)=(x-2)(x-7)(x-9)(x-12)$ how many roots is there to $f'(x)=0$ and $f''(x)=0$? without taking derivative.

What I know from Rolle's theorem that between every 2 roots of the function theres atleast 1 root of the derivative, so for example $f'(x)=0$ has atleast 3 roots.
I'm struggling to get how im supposed to know its Exactly 3 roots and not atleast without differentiating since that's what the question is asking for and I feel like im missing something.
any help is really appreciated.

Answer 1

$f(2)= f(7)=f(9)=f(12)=0$. By mean value theorem there exists atleast one $c$ in $(2,7)$ ,one $c'$ in $(7,9) $and one $ c''$ in $(9,12)$ such that $ f'(c)=f'(c')=f(c'')=0$. Also degree of $f'(x) $ is $3$ which implies $f'(x)$ has atmost 3 real roots. Therefore number of real roots of $ f'(x)$ is $3$. Let the roots of $f'(x) $ be $a,b,d$. As the roots of $f'(x)$ lie in distinct intervals( i.e $(2,7), (7,9), (9,12)$) so they are distinct. As $f'(a)= f'(b)= f'(d)=0$, again using mean value theorem and proceeding similarly we get $f''(x) $ has 2 real roots.

Answer 2

$p(x)$ has $4$ real roots.

$ x<2 :$ $ p(x) >0$;

$x \in (2,7);$ $p(x) <0;$

$x \in (7,9);$ $p(x)>0;$

$x \in (9,12):$ $p(x)<0;$

$x>12;$ $p(x)>0;$

$p(x)$ has a rel. minimum $\in (2,7);$ a rel, maximum $\in (7,9);$ a relative minimum $\in (9,12)$.

$p'(x)$ has degree $3$ and has $3$ real roots in between the zeroes of $p(x)$ (Rolle).

$p''(x)$ has degree $2$ and has $2$ real roots in between the zeroes of $p'(x)$(Rolle) .

Answer 3

The signs of $f$ on $(-\infty,\,2),\,(2,\,7),\,(7,\,9),\,(9,\,12),\,(12,\,\infty)$ are $+-+-+$, so there are three turning points, i.e. three roots of $f^\prime=0$. Similarly, $f^\prime$'s left-to-right signs progression is $-+-+$, so between its three roots are two roots of $f^{\prime\prime}=0$. This may be easier to visualize if you sketch $f$ against $x$, which will look like this.