Definef (x) =(x2 if x is rational,0 if x is irrational.Show that f is differentiable at 0 and f ′(0) = 0

To show that the function f is differentiable at 0, we need to show that the limit as x approaches 0 of [f(x) - f(0)] / x exists.

First, let's note that f(0) = 0^2 = 0.

If x is rational, f(x) = x^2. So, [f(x) - f(0)] / x = [x^2 - 0] / x = x.

If x is irrational, f(x) = 0. So, [f(x) - f(0)] / x = [0 - 0] / x = 0.

Now, let's take the limit as x approaches 0.

For rational x, the limit as x approaches 0 of x is 0.

For irrational x, the limit as x approaches 0 of 0 is 0.

Since these two limits are equal, the limit as x approaches 0 of [f(x) - f(0)] / x exists and is equal to 0. Therefore, f is differentiable at 0.

To find f ′(0), we use the definition of the derivative: f ′(0) = lim (x->0) [f(x) - f(0)] / x. As we've already found, this limit is 0. Therefore, f ′(0) = 0.

To show that the function f is differentiable at 0, we need to show that the limit as x approaches 0 of [f(x) - f(0)] / x exists.

First, let's note that f(0) = 0^2 = 0.

If x is rational, then f(x) = x^2. So, [f(x) - f(0)] / x = [x^2 - 0] / x =