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

Question

Elijah Abram · Accepted Answer

Not necessarily.

The function #f(x) = abs(x^2-4x+3)# is continuous everywhere, including #[1,3]#, but is not differentiable at #1# or at #3#. graph{abs(x^2-4x+3) [-1.183, 6.613, -2.197, 1.7]}

Paisley Johnson · Answer

undefined 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.