Question

Suppose U ⊂ Rⁿ is open, p ∈ U, and f : U → R, g : U → R, h : U → R are functions such that f(p) = g(p) = h(p), f and h are differentiable at p, f '(p) = h'(p), and f(x) ≤ g(x) ≤ h(x) for all x ∈ U. Show that g is differentiable at p and g'(p) = f '(p) = h'(p).

Answer 1

We need to show that the function g is differentiable at point p and its derivative at p, g'(p), is equal to f'(p) and h'(p).

Step-by-step explanation:

We need to show that the function g is differentiable at point p and its derivative at p, g'(p), is equal to f'(p) and h'(p).

Since f and h are differentiable at p, we know that the limit of the difference quotient of f and h as x approaches p exists. This limit gives us the derivative of f and h at p, which we know are equal.

Using the given inequality f(x) ≤ g(x) ≤ h(x), we can use the squeeze theorem to show that g is also differentiable at p and its derivative at p is equal to f'(p) and h'(p). This is because the squeeze theorem states that if a function f is squeezed between two other functions g and h, and g and h have the same limit as x approaches a, then f also has that limit at x approaches a.