Mean Value theorem question

[Jared123]

Have I answered this question correctly and have I added the right sign (-/+) for the answer?

 

[Deleted member 4993]

Have I answered this question correctly and have I added the right sign (-/+) for the answer?
View attachment 19475
View attachment 19476

I have not solved the problem myself - neither did I go through your solution critically.

However, it looks good!

Why are you doubting yourself?

 

[Jared123]

I have not solved the problem myself - neither did I go through your solution critically.

However, it looks good!

Why are you doubting yourself?

I sometime make little mistakes and it takes someone else picking it up for me to notice it.

 

[Jared123]

I have not solved the problem myself - neither did I go through your solution critically.

However, it looks good!

Why are you doubting yourself?

Also since the function is not differentiable, do I need to change my answer for c or I dont have to?

 

[Jared123]

I thought that from the diagram that sqrt(36+x^2) was not differentiable

 

[Steven G]

In my opinion the main mistake you made is that you NEVER verified the conditions of the MVT! You just can't use a theorem because you were asked to, you must verify that it satisfies the conditions of the theorem. I have always taken off points on exams when students do not verify that the conditions were satisfied.

I mentioned three times (and now for a 4th time) that you must verify the conditions of the theorem. If at work you ever use the MVT and it was not applicable you might be in deep trouble. This is a very serious mistake you made.

 

[Steven G]

I thought that from the diagram that sqrt(36+x^2) was not differentiable View attachment 19481

You have a slight typo with your function. Are you saying that the function is not differentiable on (-6,4)?

 

[Steven G]

Since you squared both sides I would have written sqrt(5)/5 as 1/sqrt(5). This way the numbers would have been smaller after the squaring.

 

[Deleted member 4993]

I did not check the conditions either. Those being:

1. f(x) is continuous on the closed interval [a,b].
   
   
2. f(x) is differentiable on the open interval (a,b).
   
   To the corner - on the double........

 

[Steven G]

I did not check the conditions either. Those being:

1. f(x) is continuous on the closed interval [a,b].
   
   
2. f(x) is differentiable on the open interval (a,b).
   
   To the corner - on the double........

Yes, no joking but that is a big error!