# If a function is continuous along the interval [1,3], would it be differentiable at x=1 and x=3?

Not necessarily.

graph{abs(x^2-4x+3) [-1.183, 6.613, -2.197, 1.7]}

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If a function is continuous on the interval [1,3], it does not necessarily imply that the function is differentiable at ( x = 1 ) and ( x = 3 ). A function can be continuous at a point without being differentiable at that point. To determine differentiability at ( x = 1 ) and ( x = 3 ), we need to check if the derivative exists at those points. This involves examining the behavior of the function near those points using the limit definition of the derivative. If the derivative exists at ( x = 1 ) and ( x = 3 ), then the function is differentiable at those points. Otherwise, it is not differentiable at those points.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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