If f is continuous at x=c, then f is differentiable at x=c
True
False
nope. Consider y = |x|
the derivative does not exist at x=0
True.
To understand why this statement is true, let's first define what it means for a function to be continuous at a point. We say a function f is continuous at x = c if three conditions are satisfied:
1. The function f(c) is defined (i.e., c is in the domain of f).
2. The limit as x approaches c of f(x) exists.
3. The limit as x approaches c of f(x) is equal to f(c).
Now, let's consider what it means for a function to be differentiable at a point. We say a function f is differentiable at x = c if the derivative of f at c exists. In other words, the following limit exists:
lim┬(x→c)〖(f(x)-f(c))/(x-c)〗
If f is continuous at x = c, then the first two conditions for continuity are satisfied. This means that the limit as x approaches c of f(x) exists.
To prove that f is differentiable at x = c, we need to show that the third condition for continuity implies the existence of the derivative limit. This can be done using the concept of differentiability based on limits.
By using the definition of the derivative, we can rewrite the limit as x approaches c of (f(x) - f(c))/(x - c) as:
lim┬(x→c)〖(f(x)-f(c))/(x-c)〗 = lim┬(x→c)(f(x)-f(c))/(x-c) × (x-c)/(x-c)
Since f is continuous at x = c, we know that lim┬(x→c)(f(x)-f(c))/(x-c) exists. Additionally, the limit of (x-c)/(x-c) as x approaches c is 1.
Therefore, by the limit multiplication property, we can conclude that the limit as x approaches c of (f(x) - f(c))/(x - c) exists. This implies that f is differentiable at x = c.
Hence, if f is continuous at x = c, it follows that f is differentiable at x = c.