Use the MVT to prove: if f′(x)=0f'(x) = 0f′(x)=0 for all x∈(a,b)x \in (a,b)x∈(a,b), then fff is constant on [a,b][a,b][a,b]. The key step is:
For any x1,x2∈[a,b]x_1, x_2 \in [a,b]x1,x2∈[a,b], MVT gives f(x2)−f(x1)=f′(c)(x2−x1)=0f(x_2)-f(x_1) = f'(c)(x_2-x_1) = 0f(x2)−f(x1)=f′(c)(x2−x1)=0
If f′=0f' = 0f′=0 everywhere, then f=0f = 0f=0 (not just constant)
Use the Fundamental Theorem of Calculus: f(b)−f(a)=∫ab0 dx=0f(b)-f(a) = \int_a^b 0\,dx = 0f(b)−f(a)=∫ab0dx=0
Both (a) and (c) give valid proofs