Tom K. answered 07/31/20
Knowledgeable and Friendly Math and Statistics Tutor
The easiest way to disprove is with counterexamples.
a) False; let f(x) = 1 x ≠3 and f(3) = -1.
This function has f(1) > 0, f(3) < 0, and no c for which f(c) = 0. The requirement missing for f(c) = 0 for some x between 1 and 3 is that f(x) is not continuous.
b)False; let f(x) = |x|
This function is continuous everywhere, but it is not differentiable at x = 0