SOLUTION: disciss continuity and differentibility at x=1 f(x)={ x if 0<=x<=1 } 2x_1 if 1<=x<=2

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) (Show Source): You can put this solution on YOUR website!

What does "2x_1" mean?

John

My calculator said it, I believe it, that settles it

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