Question

Suppose F and G are continuously∗ differentiable functions defined on [a, b] such that F0(x) = G0(x) for all x ∈ [a, b]. Using the fundamental theorem of calculus, show that F and G differ by a constant. That is, show that there exists a C∈R such that F (x) G(x) = C.

Answer 1

Answer:

The proof is detailed below.

Step-by-step explanation:

We will first prove that if H(x) is a differentiable function in [a,b] such that H'(x)=0 for all x∈[a, b] then H is constant. For this, take, x,y∈[a, b] with x<y. By the Mean Value Theorem, there exists some c∈(x,y) such that H(y)-H(x)=H'(c)(x-y). But H'(c)=0, thus H(y)-H(x)=0, that is, H(x)=H(y). Then H is a constant function, as it takes the same value in any two different points x,y.

Now for this exercise, consider H(x)=F(x)-G(x). Using differentiation rules, we have that H'(x)=(F-G)(x)'=F'(x)-G'(x)=0. Applying the previous result, F-G is a constant function, that is, there exists some constant C such that (F-G)(x)=C.