Final answer:
The given set S is a subspace of the vector space V because it satisfies the three conditions: closure under addition, closure under scalar multiplication, and the existence of the zero vector.
Step-by-step explanation:
To determine whether the set S is a subspace of the vector space V, we must check three conditions: closure under addition, closure under scalar multiplication, and the existence of the zero vector.
1. Closure under addition: Let f and g be two functions in S. Then (f+g)'(0) = f'(0) + g'(0). Since f'(0) and g'(0) are both greater than or equal to 0, their sum will also be greater than or equal to 0. Therefore, f+g is in S and S is closed under addition.
2. Closure under scalar multiplication: Let f be a function in S and c be a scalar. Then (cf)'(0) = cf'(0). Since f'(0) is greater than or equal to 0, cf'(0) will also be greater than or equal to 0. Therefore, cf is in S and S is closed under scalar multiplication.
3. Existence of the zero vector: The zero vector in this case is the function f(x) = 0. Since f'(0) = 0, which is greater than or equal to 0, the zero vector is in S.
Therefore, S satisfies all the conditions and is a subspace of V.