Understanding Differentiability of f at x = 0

[Miike012]

The paragraph says, " Even if the function f is an everywhere differentiable function, it is still possible for f ' to be discountinuous. However, the graph of f ' can never exhibit a discountinuity of ..." picture is in paint document...

What type of discountinuity is that? a hole discountinuity?

second Question: Is my understanding correct? This is my explanation why f is not differentiable at some x value.

Given
f ' (x) =
x , x>0
x^2 x<0
7 , x = 0.

Therefore, the graph of the differnetiable function should have a similar discontinuity as the one shown in the paint document. (Determining the limit)
lim f ' (x) as x → 0- = lim f ' (x) as x → 0+ = 0 however because lim Δx → 0 [ f(0 + Δx) - f(0) ]/Δx is equal to f ' (0) = 7 ≠ 0, therefore the function f non differentiable at x = 0 because inorder for f to be diff at 0, f'(0) must equal 0

Is my understanding why f is not differentiable at x = 0 correct?

Next, If I wanted to make f differentiable at x could I do this by defining one of two functions...

f ' (x) =
x , x>0
x^2 x<0
0 , x = 0.

OR
f ' (x) =
x , x>0
x^2 x<0

Would this work to make f diff at x = 0?

 

[atomthick]

The answer to your last question is yes for the first function. The second function is not defined in 0.

 

[HallsofIvy]

While the derivative of a function is not necessarily continuous, it must have the "intermediate value propert"- if f'(x0)= a and f'(x1)= then, between x0 and x1, f' must take on all values between a and b. From that, it follows that a derivative cannot have a "jump" discontinuity (the left and right limits exist but are different).