Question

Explain why it is not valid to use the Mean Value Theorem. When the hypotheses are not true, the theorem does not tell you anything about the truth of the conclusion. Find the value of c, or show that there is no value of c that makes the conclusion of the theorem true.

Answer 1

\[f(x)=x ^{1/3}, [-1,1]\]

Answer 2

What in the world is this asking me to do?

Answer 3

do u know what is the mean value theorem

Answer 4

The Mean Value Theorem is one of the most important theoretical tools in Calculus. It states that if f(x) is defined and continuous on the interval [a,b] and differentiable on (a,b), then there is at least one number c in the interval (a,b) (that is a < c < b) such that

\[f'(c) = \frac{f(b) - f(a)}{b-a} \]

Answer 5

to support huneya only. Not interfere the guiding

Answer 6

thanks i know but this problem is to plug it in the formula i did this yesterday in class that why.

Answer 7

for this particular problem f(x) = x^1/2 [-1.1]
it's NOT defined on this interval.---> the Mean theorem fails

Answer 8

yes true

Answer 9

DAT SIT

Answer 10

the answer to the question is NO, because there is a vertical tangent at x=0

Answer 11

its \[x ^{1/3}\]

Answer 12

but still follows same thing you said right, not continuous at x=0?

Answer 13

x^1/3 right than the answer when u put it the mean value is a error and than u have to justify it so u then graph it and find that there is a vertical tangent at x=0

Answer 14

If the mean value theorem does not hold for a case does that mean there is not c that satisfies the problem or do you have to do some other method to determine that?

Answer 15

the value of c is error or 0 at the interval [-1,1]

Answer 16

okay got it

Answer 17

ok

Answer 18

maybe i am doing someone wrong but i do get a value for c?

Answer 19

show me ur work

Answer 20

f(1)=1 and f(-1)=-1
f'(c)=1
f'(x)=1/3x^-2/3

set that equal to 1 and i get c equals around .19

Answer 21

ok thsi is not use the mean value theorem thats y

Answer 22

i dunno, i just followed the steps in my textbook

Answer 23

it looks like x^1/3 is continuous

Answer 24

and differentiable over all real numbers so therefore the MVT would work?

Answer 25

do u use the book calculus graphical, numerical, algebraic third edition

Answer 26

no, calculus early transcendental functions 4th edition

Answer 27

by smith and minton............. absolutely terrible book

Answer 28

i use a different book but is the same problem in my book and the answer is as i mention above

Answer 29

i dont see how you get a vertical tangent at 0, the function seems to be continuous there and the left and right sided limits agree as x goes to 0?

Answer 30

it meets all the wickets for a function to be continuous and is also differentiable in that interval so i am not seeing why the MVT does not apply

Answer 31

find the derivative of and then see why

Answer 32

gotcha, i was thinking that a value exits for that interval when plugged into the derivative. But I should have been looking at the derivative of the function over that interval, if you get what i am saying