true/ false calculus questions

Question

If f is continuous on [0 , 1] , then it is also differentiable on [0 , 1] . 

i think this is true, but. is there a particular theorem to prove this?

 

If f is differentiable on [0 , 1] , then f(x) = 0 for some x in [0 , 1] 

i think this might be true, because if it is differentiable on the interval, it must also be continuous?  i think that rolle's theorem proves this statement.  

 

If f is differentiable on [0 , 1] , and f(0) = f(1) = 6 , then f(x) = 0 for some 0 < x < 1 .

 

 

If f has no absolute maximum on [0 , 1] , then f is discontinuous.

i think this is true, but i do not know if there is a theorem that proves it? 

Answer 1

If f is continuous on [0, 1], then it is also differentiable on [0, 1].  This is true because the slope of the tangent line at each point on the interval will be different.  The mean value theorem can prove this.

 

f'(x) = 0 for some x in in [0, 1].  In this interval, there exist a maximum or minimum value, according to the mean value theorem.

 

f'(x) = 0   if f'(0) = f'(1) = 6  ,  a minimum or maximum must exist for some 0 < x < 1.

 

If f has no absolute maximum on [0, 1], then f is discontinuous.  According to the mean value theorem, if you have an open interval (0, 1), then f is discontinuous at the end boundaries since 0 and 1 are not included values.

 

 

I hope this helps you out.

Answer 2

I can help you with the first 3 questions:

 

It is important to avoid confusing continuity with being able to differentiate.  Continuity refers to the curve being connected, i.e. there are no holes or gaps in the curve.  Being able to differentiate implies "smoothness" in the shape of the curve.

 

(1) 

 

Being continuous does not guarantee it can be differentiated.  Example:

 

f(x) = x for 1 ≤ x ≤ 1/2, 

f(x) = -x + 1 for 1/2 ≤ x ≤ 1

 

Note that f(1/2) → 1/2 from either side of 1/2, but f'(1/2) is 1 approaching from the left, and -1 approaching from the right. 

 

f'(1/2-) ≠ f'(1/2+).

 

(2)

 

Counterexample:

 

f(x) = x² + 10

 

f'(x) is defined over [0,1], but f(x) ≠ 0 everywhere over [0,1]

 

(3)

 

Counterexample:

 

f(x) = 6 

 

This satisfies f(0) = f(1) = 6, but f(x) ≠ 0 everywhere over [0,1].  This example is trivial, but you could have a concave down parabola symmetric about x = 1/2, passing through (0,6) and (1,6).  It will not be zero anywhere over [0,1].

 

(4)  

 

Consider a function like 

 

f(x) = 1/(1 - x)

 

It will not be continuous at x =1 because f(x-) ≠ f(x+) as you approach x = 1, so this one looks true, but I would like to see a math tutor explain this one.