Mean Value Theorem

[layd33foxx]

Consider the function

F(x)={x² if x ϵ [0,1]
{2x-1 if x ϵ (1,2]

Idk where my mistake was please check if I'm right.

a)calculate lim f(x)-f(1)/x-1= 2
x->1+

b)calculate lim f(x)-f(1)/x-1=2

c) determine whether mean value theorem can be aplpied on [0,2]. If it can be applied then determine all value of c such that f'(c)=f(2)-f(0)/2-0.


=2(2)-1-(0)²/2-0 mean value theorem can be applied

c=3/4

 

[renegade05]

First, I just would like to clarify your question:

\(\displaystyle f(x) = \left\{
\begin{array}{lr} x^2 & :0\le x \le1\\ 2x-1 & :1<x\le2 \end{array}\)

(a) Calculate:

\(\displaystyle \lim_{x\to1^+}f(x) - \frac{f(1)}{x-1}\)

(b) Calculate:

Why do these limits equal two (because they don't)? Plus, (a) and (b) are the same question (minus the limit of course). Were you suppose to take \(\displaystyle \lim_{x\to1^-}\) in part (b) ?? Am I missing something here?

 

[layd33foxx]

Yes, you are correct.

 

[renegade05]

Yes, you are correct.

Ok so now we have:

\(\displaystyle f(x) = \left\{\begin{array}{lr} x^2 & :0\le x \le1\\ 2x-1 & :1<x\le2 \end{array}\)

Calculate:

(a) \(\displaystyle \lim_{x\to1^+}f(x) - \frac{f(1)}{x-1}\)

(b) \(\displaystyle \lim_{x\to1^-}f(x) - \frac{f(1)}{x-1}\)

(c) Determine whether mean value theorem can be aplpied on [0,2]. If it can be applied then determine all value of c such that \(\displaystyle f'(c)=\frac{f(2)-f(0)}{2-0}\)

Let's start with (a) and (b). I take it your guess at an answer is 2? That is incorrect. Want to try again? Perhaps splitting up the limit to two separate limits will help you see where this limit is going. Here it is done for part (a):

\(\displaystyle \lim_{x\to1^+}f(x) - \lim_{x\to1^+}\frac{f(1)}{x-1}\)

What is \(\displaystyle f(1)\), what is \(\displaystyle f(x)\)?

 

[lookagain]

Consider the function

F(x)={x² if x ϵ [0,1]
{2x-1 if x ϵ (1,2]

Idk where my mistake was please check if I'm right.

a)calculate lim f(x)-f(1)/x-1= 2
x->1+

\(\displaystyle Consider \ this: \ \ lim (x-> 1^+)[f(x) - f(1)]/(x - 1)\)



b)calculate lim f(x)-f(1)/x-1=2

\(\displaystyle Consider \ this: \ \ lim (x -> 1^- \ ?)[f(x) - f(1)]/(x - 1)\)


c) determine whether mean value theorem can be [applied] on [0,2].
If it can be applied then determine all value of c such that
f'(c)=f(2)-f(0)/2-0.


\(\displaystyle Consider \ this: \ \ f'(c) \ = \ [f(2) - f(0)]/(2 - 0)\)


=2(2)-1-(0)²/2-0 mean value theorem can be applied

\(\displaystyle Consider \ this: \ \ \{[2(2) - 1] - [(0)^2]\}/(2 - 0)\)


c = 3/4

layd33foxx and renegade05,

notice how the lack of necessary grouping symbols
had you typed incorrect expressions.


The alternative forms highlighted above should have been
what is meant, instead of

\(\displaystyle \lim_{x\to1^+}f(x) - \frac{f(1)}{x-1}\)


- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -


\(\displaystyle Think: \ \ \lim_{x \to 1^+} \dfrac{f(x) - f(1)}{x - 1}\)

 

[renegade05]

First, I just would like to clarify your question:

...

\(\displaystyle \lim_{x\to1^+}f(x) - \frac{f(1)}{x-1}\) ??

Yes, you are correct.

However, yes, I should have spotted the grouping error right off the bat. I don't know what I was thinking.

But, geezzz.. Did you even look at my post layd33foxx? Do you know what CLARIFY means?

 

[layd33foxx]

Um what is the big deal were you going to help me or not?

 

[renegade05]

Um what is the big deal were you going to help me or not?

You don't see the big deal in: me wanting to clarify your question, asking if the question I thought was yours but it wasn't was correct, you saying it is correct(it wasn't), and me preceding to trying to help you?

Sorry, but are you *blank* kidding me?

Do you even see what error lookagain is talking about?

 

[chowe13]

Help, i hope

The MVT can be applied if:

1. the function is continuous on the set domain [0,2]
2. the function is differentiable on the domain [0,2]

So...since this is a piece-wise function it must have the same "y" value at the point the graphs join, so substitute 1 into x in both equations and they do in fact have the same value so it is continuous. Both pieces are continuous on there own so the MVT can be used.

USing the MVT, the average(secant) slope between the two points is: 3/2

set each of the two derivatives equal to 3/2 and solve to find all c values.


The limit should be the same from both 1+ and 1-, it is equal to 1. If it were not the same you could not by definition use the MVT because it would not be continuous