Part (a)
The graph is continuous at x = 0 because we have a connected curve at this point. There are no jumps or gaps at x = 0.
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Part (b)
The graph is NOT differentiable at x = 0 because of the sharp point. There isn't a smooth transition from one tangent slope to the other. If you were to apply the derivative to y = |x|, then you'll find the graph of y' = |x|/x has a jump discontinuity at x = 0.
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Part (c)
The graph is NOT continuous at x = 1 because of the open hole here. Think of it as a pothole in the ground that you can't drive over. We consider this a removable discontinuity since we basically pulled exactly one point from the graph to toss away. The graph is continuous everywhere else but x = 1.
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Part (d)
The graph is NOT differentiable at x = 1 because we need continuity as one of the conditions. But unfortunately as mentioned in part (c), the graph isn't continuous at x = 1.
If a portion isn't continuous, then it's certainly not differentiable. Think of the "differentiable" building block dependent on the "continuous" building block as a foundation. This is because the differentiability of a function depends on a limiting value based on whether f(x) exists at that location.
As mentioned earlier, just because something is continuous, it doesn't automatically lead to differentiability. See part (b).
But luckily, if you know something is differentiable at a certain spot, then it must be continuous there as well.
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Summary of the answers:
- (a) Continuous
- (b) Not differentiable
- (c) Not continuous
- (d) Not differentiable
Let me know if you have any questions.