Write a function that reduces compositions of linear operators

GolfScript, 41 40 23 19 characters

{2,0+%/-{@*+}++}:k;

Thanks to Peter Taylor for pointing out this solution. Accepts two function bodies as arguments.

The following examples show how to use function k:

# As function
{2*3+} {3*4+} k         # => {11 6 @*+}

# Apply to value
5 {2*3+} {3*4+} k ~     # => 41

If one wants to retain the original output format, the following 22 character solution is possible:

{2,0+%/-{*}+\+{+}+}:k;

{2*3+} {3*4+} k         # => {6 * 11 +}

Explanation of the code:

{                       # on stack are functions g h
  2,                    # build array [0 1] -> g h [0 1]
  0+                    # append 0 -> g h [0 1 0]
  %                     # apply function h to each element -> g [h(0) h(1) h(0)]
  /                     # apply function g to each element
                        # and break up the array -> g(h(0)) g(h(1)) g(h(0))
  -                     # subtract last two -> g(h(0)) g(h(1))-g(h(0))
  {@*+}++               # append function body to these values
}:k;                    # save function to variable k

Mathematica 15

Define g, h as pure functions

g and h are defined as pure functions (awaiting a value or independent variable to be assigned)

g = 2 # + 3 &;
h = 3 # + 4 &;
g[x]
h[x]

3 + 2 x
4 + 3 x

---

Method #1: Compose g of h directly (15 chars)

g@h@x//Simplify

11 + 6 x

Or, alternatively: (17 chars)

g[h[x]]//Simplify

---

Method #2: Define f (25 chars= 13 + 10 chars for Simplify)

f=Composition

f[g, h][x]
f[g, h][x] // Simplify
Composition[g, h][x]//Simplify
f[g, h][1]
Table[f[g, h][x], {x, 1, 13}]

3 + 2 (4 + 3 x)
11 + 6 x
11 + 6 x
17
{17, 23, 29, 35, 41, 47, 53, 59, 65, 71, 77, 83, 89}

Haskell, 31

(f%g)x=f(g 0)+x*(f(g 1)-f(g 0))

Version in which the function body contains only variables (equivalent at run time, but closer to the question's form), 38 characters:

(f%g)x=a*x+b where h=f.g;b=h 0;a=h 1-b

Note that in both cases, since Haskell uses curried functions, there is no explicit construction of the composed function as it is implicit in % being defined with three arguments: function f, function g, value x.

Example use:

> ((\x -> 2*x + 3)%(\x -> 3*x + 4)) 0
11
> ((\x -> 2*x + 3)%(\x -> 3*x + 4)) 4
35

Ungolfed second form:

(f % g) x = a * x + b        -- equivalent: (f % g) = \x -> a * x + b
  where
    h = f . g                -- use built in function composition to bootstrap
    b = h 0                  -- constant term of composed function
    a = h 1 - h 0            -- linear term of composed function