AP Calculus BC: Differentiability and continuity

[CrazyNeutrino]

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.

 

[BvU]

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 ?

 

[Dick]

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?

 

[BvU]

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

 

[CrazyNeutrino]

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

 

[CrazyNeutrino]

Also, I'm a bit lost as to how i would test the first condition, Inorder to do so wouldn't i need to know that h(x) lies mostly or wholly above the x axis?

 

[Ray Vickson]

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.

 

[JonnyG]

Also, I'm a bit lost as to how i would test the first condition, Inorder to do so wouldn't 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.

 

[JonnyG]

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.

 

[BvU]

What i get is h'(x) = h'(2-x) (-1) = -h'(2-x)

 

[CrazyNeutrino]

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)?

 

[JonnyG]

Yes. Now plug in x = 1. What do you get?

 

[CrazyNeutrino]

h'(1)= -h'(1) ?
How does this make sense?

 

[JonnyG]

Good. Now solve for h'(1)

 

[CrazyNeutrino]

Unless this always implies that h'(x) = 0? Is this reasoning correct?

 

[JonnyG]

It doesn't imply that h'(x) = 0. It implies h'(1) = 0. Can you show me why?

 

[CrazyNeutrino]

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

 

[CrazyNeutrino]

Okay so h'(1)=-h'(1)

2h'(1) = 0

h'(1) = 0

Is this correct?

 

[JonnyG]

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.

 

[CrazyNeutrino]

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?

 

[JonnyG]

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?

 

[CrazyNeutrino]

Any negative number? Say h(x) = -1
If so, aren't we assuming the value of h(x)? How can we say for sure

 

[JonnyG]

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.

 

[JonnyG]

If so, aren't we assuming the value of h(x)? How can we say for sure

We are giving a counter example.

 

[CrazyNeutrino]

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?

 

[JonnyG]

You are defining a function h whose value h(x) is equal to -1 for every real number x. Since h(x) = -1 for every x, then surely h(x) = h(2-x).

 

[CrazyNeutrino]

Ohh so if I'm not mistaken, we are simply trying to find a possible function that doesn't satisfy 1 &3 but could possibly satisfy h(x)=h(2-x)

Thank you so much for your time and help! :)

 

[JonnyG]

More specifically, you have already proven that if a function is differentiable and h(x) =h(2-x) then condition 2 is automatically satisfied. But now you have provided a function that is differentiable and h(x) =h(2-x), but violates condition 1 and 3. So you can only conclude condition 2 must be satisfied.

 

[Vijayakrishna]

It looks like h(x)=h(2-x) is possible only it is a linear function of the form y=b (a horizontal line). Can not think of any other function that satisfies this condition. As it is horizontal line the first derivative is 0 for all x. Hence II is true and as a consequence III is false. If b value is negative the definite integral will be negative hence I is false.

 

[pasmith]

It looks like h(x)=h(2-x) is possible only it is a linear function of the form y=b (a horizontal line). Can not think of any other function that satisfies this condition.

Think again. $$h(x) = (x - 1)^2 = (1 - x)^2 = (2 - x - 1)^2 = h(2 - x)$$ is one example. In general we must have $h(x) = f(x-1)$ where $f$ is any even function ($f(-x) = f(x)$).