Problem 2. Prove that the function f : R → R, f (x) = |x|3 is twice differentiable at anypoint a ∈ R, but is not three-times differentiable at 0.

Sure, let's break this down:

Step 1: Define the function and its derivatives

The function is f(x) = |x|^3.

The first derivative of f(x) is f'(x) = 3x|x|.

The second derivative of f(x) is f''(x) = 6|x|.

Step 2: Prove that f(x) is twice differentiable at any point a ∈ R

To prove that f(x) is twice differentiable at any point a ∈ R, we need to show that the second derivative exists for all real numbers.

For x ≠ 0, the second derivative f''(x) = 6|x| is clearly defined.

At x = 0, we need to check the limit of the first derivative as x approaches 0 from the left and right:

lim (x->0+) f'(x) = lim (x->0+) 3x|x| = 0

lim (x->0-) f'(x) = lim (x->0-) 3x|x| = 0

Since these two limits are equal, the derivative at x = 0 exists and is equal to 0. Therefore, f(x) is twice differentiable at any point a ∈ R.

Step 3: Prove that f(x) is not three-times differentiable at 0

The third derivative of f(x) is f'''(x) = 6 sign(x), where sign(x) is the sign function.

At x = 0, the third derivative is undefined because the sign function is not defined at x = 0. Therefore, f(x) is not three-times differentiable at 0.