Define g : R → R byg(x) =(x2 cos(1/x) if x̸ = 0,0 if x = 0.Show that g is differentiable at x for every x ∈ R

To show that the function g is differentiable at x for every x ∈ R, we need to show that the derivative of g exists at every point in R.

The function g is defined as g(x) = x²cos(1/x) for x ≠ 0 and g(x) = 0 for x = 0.

1. For x ≠ 0, the function g is a product of two functions: h(x) = x² and f(x) = cos(1/x). Both of these functions are differentiable for x ≠ 0. The derivative of a product of two functions is given by the product rule: (f*g)' = f'g + fg'. So, we can find the derivative of g for x ≠ 0 using the product rule.

2. For x = 0, we need to show that the limit as x approaches 0 of [g(x) - g(0)] / (x - 0) exists. This is the definition of the derivative at a point.

   g(x) - g(0) = x²cos(1/x) - 0 = x²cos(1/x).

   So, we need to find the limit as x approaches 0 of [x²cos(1/x)] / x.

   This limit can be evaluated using L'Hopital's rule, which states that the limit as x approaches a of [f(x) - f(a)] / (x - a) is equal to the limit as x approaches a of f'(x) / g'(x), where f and g are differentiable functions and g'(a) ≠ 0.

   Using L'Hopital's rule, we find that the limit as x approaches 0 of [x²cos(1/x)] / x is equal to the limit as x approaches 0 of [2xcos(1/x) + x²sin(1/x)(-1/x²)] / 1, which simplifies to the limit as x approaches 0 of [2cos(1/x) - sin(1/x)].

   Both cos(1/x) and sin(1/x) are bounded functions, meaning they always lie between -1 and 1. Therefore, as x approaches 0, [2cos(1/x) - sin(1/x)] also lies between -3 and 3.

   Therefore, the limit as x approaches 0 of [g(x) - g(0)] / (x - 0) exists, and so the derivative of g at x = 0 exists.

Therefore, the function g is differentiable at x for every x ∈ R.