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?