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Miike012
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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?
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?
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