#calc
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The graph of f is "broken" at x = 1
So f is not differentiable at x = 1, don't worry about B and C
The graph looks smooth around x=0, I suppose f''(x) is also continuous at 0, so there is only one suitable value bridging the negative values of f''(x) when -1 < x < 0 and the positive values of f''(x) when 0 < x < 1
Similarly, the graph is sloping upwards to the right of x=2; this should tell us the sign of f'(x) there
Continuity of f'' may not be necessary, since the intermediate value property is guaranteed by Darboux
am i wrong..?