If the function $$f(x)=x^3-9x^2+24x+c$$ has three real and distinct roots $l,m,n$, the find value of $[l]+[m]+[n]$ where $[..]$ represents greatest integer function
In my book there is no information given about $c$.
First you can establish that the graph must have stationary points at $x=2$ and $x=4$
For there to be three real distinct roots, you require the values of $f(2)$ and $f(4)$ to be of opposite parity.
This gives rise to the condition that $-20<c<-16$
For each of the extreme values of $c$ the polynomial factorises nicely and from a sketch you will be able to deduce the quantity you are looking for.
I hope this helps
The discriminant of this cubic is $-27 c^2 - 972 c - 8640 = -27(c+20)(c+16)$. To have three distinct real roots, we need the discriminant positive, so $-20 < c < -16$.
Note that the values of $f$ at $x = 1,2,3,4,5$ are $c+16, c+20, c+18,c+16,c+20$ respectively.
For $c$ between $-20$ and $-16$, the lowest root is between $1$ and $2$, the highest between $4$ and $5$. The middle root is between $2$ and $4$, is an increasing function of $c$, and is $3$ when $c=-18$. So for $-20 < c < -18$ we have $\lfloor l \rfloor + \lfloor m\rfloor + \lfloor n\rfloor = 1+2+4=7$, and for $-18 \le c < -16$ we have $1+3+4 = 8$.
From Vieta's formula:
If $l,m,n\in\mathbb R$, then
$$\lfloor l\rfloor+\lfloor m\rfloor+\lfloor n\rfloor\ge l+m+n=9$$
Therefore, we have
$$6\le\lfloor l\rfloor+\lfloor m\rfloor+\lfloor n\rfloor\le9$$
Since we have local maximum and minimum at $x=2,4$, then