TITLE: What is the "universal problem" that motivates the definition of homotopy limits/colimits (and more generally "derived" functors)?
QUESTION [12 upvotes]: The ordinary notions of limit and colimit are universal solutions to a problem, specifically, finding terminal/initial objects in slice/coslice categories.  In the context of homotopy right Kan extensions (it's not hard to show that the theory of homotopy limits reduces to this case (the same holds for left Kan extensions/colimits)), we say that given a functor $f:C\to C'$ between small categories, and a functor $F:C\to A$ where $A$ is a cofibrantly-generated model category, that a natural transformation of functors $\alpha: H\to Ran_f F$ exhibits $H$ as the homotopy right Kan extension of $F$ if there exists an injectively fibrant replacement $G$ of $F$ such that the composite $H\to Ran_fF \to Ran_fG$ is a weak equivalence.  
When $A$ is combinatorial, we can also simply define a homotopy right Kan extension functor along $f$ to be $Ran_f (Q (-))$, where $Q$ is a functorial injectively fibrant replacement functor.  
This is easy enough to define, but why is this the definition?  Why would we want to take fibrant/cofibrant replacements and consider their ordinary Kan extensions/limits/etc?  I suspect that it has to do with the fact that homotopy is an honest equivalence relation on arrows from a cofibrant object into a fibrant object, but I would appreciate an actual explanation.

REPLY [4 votes]: This could have been a comment to Peter's answer, except that it's going to be long. (Maybe I should be making it a new question.)
Suppose that $Ho(\mathcal C)$ is obtained from $\mathcal C$ by (universally) inverting a class $\mathcal W$ of morphisms. Let's call an object of $\mathcal C$ homotopically terminal if in $Ho(\mathcal C)$ it is terminal. 
I did not assume anything about $\mathcal W$ (except that this localization existed). In practice we usually have more to work with, like a model structure or at least the 2 out of 6 condition. Given enough structure or information, there are ways of defining a simplicial set for each pair of objects such that this serves as a "function space", in particular such that its set of components is the set of morphisms in $Ho(\mathcal C)$. I'm not a master of the known sufficient conditions for making these function spaces -- I suppose that there are several overlapping approaches. 
QUESTION: Can we say that (in any or all of these approaches) an object $X$ is homotopically terminal in the sense I gave above if and only if for every object $Y$ the function space $Hom(Y,X)$ is weakly equivalent to a point? I hope so.
Here is a general, relatively simple, approach to the idea of derived functor, really just a distillation of standard ideas as I understand them, but trying to steer away from extraneous optional technicalities:
Suppose that in $\mathcal C$ and in $\mathcal D$ there are classes of morphisms $\mathcal U$ and $\mathcal V$ respectively. Let $\mathcal F$ be the category of functors $\mathcal C\to \mathcal D$. In $\mathcal F$ let $h\mathcal F$ be the full subcategory of "homotopy-invariant" functors -- those which take $\mathcal U$ into $\mathcal V$. By a left derived functor of $F\in \mathcal F$ I mean the following: Form the category $hF$ of all homotopy-invariant functors over $F$. An object is a pair $(G,G\to F)$ where $G\in h\mathcal F$. A morphism is a map $G_1\to G_2$ compatible with the maps to $F$. Define the class of maps $\mathcal W$ in $hF$ by requiring that for every $X\in \mathcal C$ the map $G_1(X)\to G_2(X)$ belongs to $\mathcal V$. By a left derived funtor of $F$ let us mean a homotopically terminal object of the category $hF$ (with respect to this class $\mathcal W$).
The standard way to achieve such a thing is to use a "replacement functor" $Q:\mathcal C\to \mathcal C$. You need a map $Q\to 1$, and you need three things to be true: (1) $QX\to X$ always belongs to $\mathcal U$. (2) $Q$ takes $\mathcal U$ into $\mathcal U$. (3) As applied to maps between objects that are in the image of $Q$, $F$ takes $\mathcal U$ into $\mathcal V$. I believe that that's all you need to show that $F\circ Q$ is homotopically terminal in the sense I defined. The proof is basically contained in what Peter wrote, including the edits prompted by my comments.
You don't even need the 2 out of 3 condition for $\mathcal U$ or $\mathcal V$ to get this. 
Right?