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AP Calculus BC: Differentiability and continuity

  • #1

Homework Statement


The function h is differentiable, and for all values of x, h(x)=h(2-x) Which of the following statements must be true?

1. Integral (from 0 to 2) h(x) dx >0
2. h'(1)=0
3.h'(0)=h'(2)=1

A. 1 only
B.2 only
C. 3 only
D. 2 &3 only
E. 1,2 &3

Homework Equations



None that I am aware of

The Attempt at a Solution


If h(x) = h(2-x) then h'(x) must also = h'(2-x)
Therefore, when x=0, h'(0)=h'(2)
However, we cannot say for sure that both = 1. Thus I eliminate any options with '3'.
Have been stuck here for a while, please help.
 

Answers and Replies

  • #2
BvU
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If h(x) = h(2-x) then h'(x) must also = h'(2-x)
What if you look at statement 2 with this conclusion in mind ?
 
  • #3
Dick
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Homework Statement


The function h is differentiable, and for all values of x, h(x)=h(2-x) Which of the following statements must be true?

1. Integral (from 0 to 2) h(x) dx >0
2. h'(1)=0
3.h'(0)=h'(2)=1

A. 1 only
B.2 only
C. 3 only
D. 2 &3 only
E. 1,2 &3

Homework Equations



None that I am aware of

The Attempt at a Solution


If h(x) = h(2-x) then h'(x) must also = h'(2-x)
Therefore, when x=0, h'(0)=h'(2)
However, we cannot say for sure that both = 1. Thus I eliminate any options with '3'.
Have been stuck here for a while, please help.
I'm having a hard time agreeing with the statement that ##h'(x)=h'(2-x)##. Why don't you try and construct a function that satisfies ##h(x)=h(2-x)## and differentiate it? Then substitute ##2-x## for ##x## and see what happens to the derivative?
 
  • #4
BvU
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Agree with DIck. My mistake. Re-think ##h'(x) = h'(2-x)## (e.g using the chain rule) That way you get something much more usable for x = 1 :smile:
 
  • #5
I tried using the chain rule. What i get is h'(x) = h'(2-x) (-1) = -h'(2-x)
When evaluated at x=1 thos gives h(1)= -h(1) which makes no sense to me
 
  • #6
Also, I'm a bit lost as to how i would test the first condition, Inorder to do so wouldnt i need to know that h(x) lies mostly or wholly above the x axis?
 
  • #7
Ray Vickson
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I tried using the chain rule. What i get is h'(x) = h'(2-x) (-1) = -h'(2-x)
When evaluated at x=1 thos gives h(1)= -h(1) which makes no sense to me
No, that is not what that equation gives. Try again.
 
  • #8
181
17
Also, I'm a bit lost as to how i would test the first condition, Inorder to do so wouldnt i need to know that h(x) lies mostly or wholly above the x axis?
There is an easy counter-example to rule out the first and third condition.
 
  • #9
181
17
I tried using the chain rule. What i get is h'(x) = h'(2-x) (-1) = -h'(2-x)
When evaluated at x=1 thos gives h(1)= -h(1) which makes no sense to me
You don't get h(1) = -h(1) from the previous equation. Be careful.
 
  • #10
BvU
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  • #11
I'm confused. According to the chain rule its the derivative of the outer function X the derivative of the inner function. Doesn't that give h'(x) = h'(2-x)(-1)?
 
  • #12
181
17
Yes. Now plug in x = 1. What do you get?
 
  • #13
h'(1)= -h'(1) ?
How does this make sense?
 
  • #14
181
17
Good. Now solve for h'(1)
 
  • #15
Unless this always implies that h'(x) = 0? Is this reasoning correct?
 
  • #16
181
17
It doesn't imply that h'(x) = 0. It implies h'(1) = 0. Can you show me why?
 
  • #17
Phew, alright i think i get that part. Consequentially I understand we can eliminate condition 3 because h'0 will be -h'(2) and not h'(2) and also there is no way to say they are equal to 1 (Hope I'm right). How do you deal with the first condition though? I see no way to test the value of the integral
 
  • #18
Okay so h'(1)=-h'(1)

2h'(1) = 0

h'(1) = 0

Is this correct?
 
  • #19
181
17
Okay so h'(1)=-h'(1)

2h'(1) = 0

h'(1) = 0

Is this correct?
This is correct. You can eliminate conditions 1 and 3 by using a counter example. There is a certain function, whose derivative is zero, that you can use. It satisfies condition 2 but contradicts condition 1 and 3.
 
  • #20
I'm afraid I don't follow. Wouldn't it have to be a constant function for the derivative to be zero? Why is important that the derivative of this counter example must be zero?
 
  • #21
181
17
I'm afraid I don't follow. Wouldn't it have to be a constant function for the derivative to be zero? Why is important that the derivative of this counter example must be zero?
Yes it has to be a constant function. But which constant function exactly would you need in order for condition 1 to be contradicted?
 
  • #22
Any negative number? Say h(x) = -1
If so, aren't we assuming the value of h(x)? How can we say for sure
 
  • #23
181
17
That function works.

So that function is differentiable and satisfies h(x) = h(2-x), right? It also satisfies condition 2 (which you already proved it must). But it contradicts condition 1 and 3. So you can conclude that given a differentiable function h(x) such that h(x) = h(2-x), only condition 2 must be true.
 
  • #24
181
17
If so, aren't we assuming the value of h(x)? How can we say for sure
We are giving a counter example.
 
  • #25
Alright i follow that logic except for how h(x) = -1 satisfies h(x)=h(2-x)

If h(x) = -1 how do we know this implies h(2-x)= -1

Doesn't the condition hold for only a specific function and not any i choose?
 

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