SOLUTION: disciss continuity and differentibility at x=1 f(x)={ x if 0<=x<=1 } 2x_1 if 1<=x<=2
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Question 1022424
:
disciss continuity and differentibility at x=1
f(x)={ x if 0<=x<=1 }
2x_1 if 1<=x<=2
Answer by
solver91311(24713)
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What does "2x_1" mean?
John
My calculator said it, I believe it, that settles it