Let f(x) = x/2 if x is even, f(x) = 3x+1 if x is odd. For any two natural numbers x, y, define the relation ">" by: x > y iff f^n(x) = y for some positive n, where f^n means n iterations of f.
Let C(x) be the function defined in HW 10, problem 3. Note that if x > y, then either C(x) and C(y) are both defined or both undefined; and if they're both defined, then C(x) > C(y).
Thus, in the recursion relation D(x) = 1 + D(f(x)), we see that D(x) is defined in terms of "its previous value" D(f(x)) (since x > f(x) according to the above definition). Furthermore, if there are only finitely many numbers x_i such that x > x_1 > x_2 > ... (i.e., if C(x) is defined), then D(x) can be computed, while if there are infinitely many such x_i (i.e., C(x) is undefined), then D(x) is also undefined.
For a rigorous proof (though not required for this problem): Step 1. Use induction to show that for all x, D(x) = C(x). Induction Step: Assume that for some n > 0, D(x) = C(x) for every x satisfying C(x) = n; show the same is true for n+1. Step 2. Since we've already shown C(x) is computable, it follows that D(x) is too.