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

The function g is defined as g(x) = x^2 cos(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) = x^2 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 + f*g'. 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 exists. This is the definition of the derivative at a point.

g(x) - g(0) = x^2 cos(1/x) - 0 = x^2 cos(1/x).

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

We can use the squeeze theorem to show that this limit is 0. The function x cos(1/x) is always between -x and x, and the limit as x approaches 0 of -x and x is 0. Therefore, by the squeeze theorem, the limit as x approaches 0 of x cos(1/x) is also 0.

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

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

The function g is defined as g(x) = x^2 cos(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) = x^2 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 + f*g'. 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 exists. This is the definition of the derivative at a point.

g(x) - g(0) = x^2 cos(1/x) - 0 = x^2 cos(1/x).

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

We can use the squeeze theorem to show that this limit is 0. The function x cos(1/x) is always between -x and x, and the limit as x approaches 0 of -x and x is 0. Therefore, by the squeeze theorem, the limit as x approaches 0 of x cos(1/x) is also 0.

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

Knowee AI · Accepted Answer

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

The function g is defined as g(x) = x^2 cos(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) = x^2 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 + f*g'. 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 exists. This is the definition of the derivative at a point.

g(x) - g(0) = x^2 cos(1/x) - 0 = x^2 cos(1/x).

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

We can use the squeeze theorem to show that this limit is 0. The function x cos(1/x) is always between -x and x, and the limit as x approaches 0 of -x and x is 0. Therefore, by the squeeze theorem, the limit as x approaches 0 of x cos(1/x) is also 0.

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

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

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