Consider the function f(x) = piecewise [(x^3)(cos(1/x)) , x=/=0], [0, x=0]. Use the definition of the derivative at a point to determine if f is differentiable at x=0
I am not sure how to start this question. Am I supposed to plug in x = 0?
To determine if a function is differentiable at a certain point, we need to check if the derivative exists at that point. The definition of the derivative at a point is:
f'(a) = lim(h->0) [f(a + h) - f(a)] / h
For this problem, we need to calculate the derivative of f(x) at x = 0.
Let's start by calculating f(0 + h) and f(0) using the piecewise definition of the function:
f(0 + h) = (0 + h)^3 * cos(1/(0 + h)) = h^3 * cos(1/h)
f(0) = 0
Now, let's substitute these values into the definition of the derivative:
f'(0) = lim(h->0) [f(0 + h) - f(0)] / h
= lim(h->0) [h^3 * cos(1/h) - 0] / h
= lim(h->0) [h^3 * cos(1/h)] / h
= lim(h->0) h^2 * cos(1/h)
To evaluate this limit, we need to analyze the behavior of cos(1/h) as h approaches 0.
Since cos(x) is a bounded function (its values range between -1 and 1), we can use the squeeze theorem to determine the limit:
-1 <= cos(1/h) <= 1
Multiplying both sides by h^2:
- h^2 <= h^2 * cos(1/h) <= h^2
As h approaches 0, we have:
lim(h->0) - h^2 = 0
lim(h->0) h^2 = 0
Using the squeeze theorem, we conclude:
lim(h->0) h^2 * cos(1/h) = 0
Therefore, according to the definition of the derivative, f'(0) = 0.
Since the derivative f'(0) exists, we can conclude that f(x) is differentiable at x = 0.