Question

Rolle's Theorem and Mean Value Theorem

Answer 1

very nice theorems those

Answer 2

sorry had to upload the graph, I just need someone to double check my thinking here

Answer 3

Rolle's Thm:
I have to check 3 conditions:
1) Is the function continuous on [a,b]
2) Is the function differentiable on (a,b)
3) f(a)=f(b) ?

Answer 4

Problem 19 fails Rolle's Theorem because it is not continuous on [a,b] at x=-3
correct?

Answer 5

so for problem 19, Rolle's Theorem does not apply.

Answer 6

For problem 20, Rolle's Theorem does not apply because function is not differentiable at x=-1 (ie a cusp - I believe)

Answer 7

right

Answer 8

Am I correct for both?

Answer 9

yes nothing applies on that interval

Answer 10

ok now Mean Value Thm. Let me go review that theorem, need a moment

Answer 11

ok I found \[\frac{ f(b)-f(a) }{ b-a }=\frac{ 1 }{ 3 }\]

Answer 12

not sure what f'(c) is?

Answer 13

The MVT says there's a \(c\) such that \(f'(c)\) is equal to the slope you found. So \(f'(c)=\dfrac{1}{3}\).

Answer 14

but am I to find the value of c?

Answer 15

Correct. The pdf isn't loading properly for me, so I can't make out all the details, but ideally you're given a function \(f(x)\). You would take the derivative and find \(f'(x)\), then evaluate it at \(c\). Then you can solve for \(c\) by setting it equal to \(\dfrac{1}{3}\).

Answer 16

I was given a graph of a piecewise function

Answer 17

Right, I see the plot. Hmm... if you're not given an explicit form for \(f(x)\), you should be able to make one from the picture. As near as I can tell, the function is something like
\[f(x)=\begin{cases}(x+3)^2+5&\text{for }x<-3,~-3<x\le-1\\3&\text{for }x=-3\\
\sqrt{x+1}+1&\text{for }x>-1\end{cases}\]

Answer 18

yes you are correct about that function I was just about to type it

Answer 19

well I can just take the derivative of it. I am just afraid that the test will not give me a nice function and just a graph

Answer 20

Any derivatives you take won't be too bad, but you have to be a bit careful about taking the derivative of a piecewise function like this one. The derivative won't exist at points where the function is discontinuous.

Answer 21

ok c is 5/4

Answer 22

ok I checked it and is it true. Good thing we have graphing calculators......

Answer 23

and Open Study

Answer 24

Let's see... over [-1,8], \(f(x)=\sqrt{x+1}+1\), then \(f'(x)=\dfrac{1}{2\sqrt{x+1}}\).
\[f'(c)=\frac{1}{2\sqrt{c+1}}=\frac{1}{3}~~\Rightarrow~~c=\frac{5}{4}\]
It's correct!

Answer 25

Thanks, that is what I did on my paper. Sorry, I am not too well verse with Latex or too fond of using the equation editor on this site.

Answer 26

You're welcome. I'm glad the site support Latex in the first place

Answer 27

one day I will learn Latex.......one day