f(x) = x^3 - 2x, x<=2
x + 2, x>2
Why do they say f'(2) is undefined?
TIA
I a f' = 3x^2-2 x<=2 and f'(2)=10
f' = 1 x>2 f'(2+)= 1
So although f(x) is continous at x=2, the curve f' is not. Therefore, f'(2) is undefined. Remember the basic definition of f' in limits, from the left, and right.
The reason why it is said that f'(2) is undefined is because the derivative of a function is not always defined at every point. In this case, let's consider the function f(x) = x^3 - 2x, x <= 2 and f(x) = x + 2, x > 2.
To find the derivative of f(x), we need to consider the derivative separately for x <= 2 and x > 2.
For x <= 2:
f'(x) = 3x^2 - 2
For x > 2:
f'(x) = 1
So you correctly calculated that f'(2) = 10 for x <= 2 and f'(2+) = 1 for x > 2. However, it is important to note that in the case of f(x), even though the function is continuous at x = 2, the curve of its derivative f'(x) is not continuous at x = 2. This means that the derivative does not have a well-defined value at x = 2.
In terms of limits, you can think of it as approaching x = 2 from the left side (x < 2), where the derivative is 3x^2 - 2, and approaching x = 2 from the right side (x > 2), where the derivative is 1. Since these two limits are different, the derivative does not exist or is undefined at x = 2.
So in conclusion, f'(2) is undefined because the curve of the derivative is not continuous at x = 2, even though the original function f(x) is continuous at that point.