Proving Interval Inclusion & Differentiability of f: I→R

[rainwyz0706]

Let I be an open interval in R and let f : I → R be a differentiable function.
Let g : T → R be the function defined by g(x, y) =(f (x)−f (y))/(x-y)
1.Prove that g(T ) ⊂ f (I) ⊂ g(T ) (The last one should be the closure of g(T), but I can't type it here)
2. Show that f ′ (I) is an interval.

I don't know what the closure of g(T) is, so I can't prove the second part in question 1. To show that f ′ (I) is an interval, I intend to show that it's connected, f(x) is continuous. Would that work? Or do I need to use g(t) is continuous to show that?
Your help is greatly appreciated!

 

[lippyka]

1. To prove that g(T) ⊂ f(I), let (x, y) ∈ T such that x ≠ y and consider f(x) - f(y). By definition of g, we have g(x, y) = (f(x) - f(y))/(x-y). Since (x, y) ∈ T, we have that x, y ∈ I and hence f(x), f(y) ∈ f(I). Thus, (f(x) - f(y))/(x-y) ∈ f(I). This shows that g(T) ⊂ f(I). To prove that the closure of g(T) ⊂ f(I), let (x, y) ∈ T such that x = y. Since (x, y) ∈ T, we have that x, y ∈ I and hence f(x), f(y) ∈ f(I). Thus, the limit of (f(x) - f(y))/(x-y) as x approaches y is equal to f(x) which is in f(I). This shows that the closure of g(T) ⊂ f(I).2. To show that f′(I) is an interval, let x, y ∈ I such that x < y. We need to show that f′(x) ≤ f′(z) ≤ f′(y) for all x ≤ z ≤ y. Since f is differentiable, it follows from the Mean Value Theorem that there exists a c between x and y such that f′(c) = (f(x) - f(y))/(x-y). Let z be any point between x and y such that x ≤ z ≤ y. Using the Mean Value Theorem again, we get that there exist c1, c2 between x and z, and z and y respectively, such that f′(c1) = (f(x) - f(z))/(x-z) and f′(c2) = (f(z) - f(y))/(z-y). Thus, since c1, c2 are between x and z, and z and y respectively, it follows that f′