<s>[INST] What is the value of $N$? $N$ is the smallest positive integer such that for any integer $n > N$, the quantity $n^3 – 7n^2 + 11n – 5$ is positive. What is the value of $N$?
Note: $N$ is a single digit number.
 [/INST] 
$\text{We have}\quad f(n)=n^3-7n^2+11n-5\tag{1}.$ 

$\textrm{We want to solve for}\;\; n\;\textrm{ when}\;\;f(n) = n^3 - 7n^2 + 11n - 5 = 0\tag{2}$  
Note that replacing $n$ with $1$ in equation $(2)$, and then summing, gives us: 
$$1(1^3) - 7(1^2 + 11(1) - 5 =  1 - 7 + 11 - 5 = 0,$$ and so $n = 1$ is a solution to $(2)$, and so $(n - 1)$ is at least one factor of the polynomial $(1)$. 
Dividing $\;f(n) = n^3-7n^2+11n-5\;$ by $\;(n - 1)$ gives us $f(n)/(n-1) = (n^2 - 6n + 5).$ 
We are then able to factor $(1)$:
$$f(n)=(n-1)(n^2-6n+5)=(n-1)[(n-1)(n-5)]=(n-1)^2(n-5).\tag{3}$$
Now we look at the roots of $f(n)$; that is, when is it true that $f(n) = (n-1)^2(n-5) = 0\;$?. 
 - We already know that when $n = 1$, $f(n) = 0$. 
 - It is also the case that when $n = 5$, $f(n) = 0$. 
So what must the value of $n$ be if we must have $f(n) > 0$?. 


*

*This is true when $n < 1$, but then, since $n$ is an integer, $n = 0$ is the greatest integer less than $1$, and we know that $n > N$, meaning $N \le -1$. But we need a positive $N$. 

*When $1 < n < 5$, $f(n) < 0.$ No good. 

*When $n > 5$, $f(n) > 0$. So we must have $n = 6$. This is the smallest positive integer for which $f(n)>0$. 


Key Note: Now, knowing $n$ is NOT ENOUGH! This $n$ is NOT the answer to your question.


*

*We need to use $n$ to determine the integer $N$. We want the smallest positive integer $N$ such that for any $n > N$, $f(n)$ is positive.

*When $N = 5$, we have that for all $n > 5$, $f(n) > 0$: this is your answer to this question.

</s>