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Mathematics 18 Online
OpenStudy (unknownrandom):

Find c using the Mean Value Theorem.

OpenStudy (unknownrandom):

Decide whether the function f=1+1/x satisfies the hypotheses of the MVP on the interval [a,b]=[1,4].

OpenStudy (unknownrandom):

So f(4)=5/4 and f(1)=2 ((5/4)-2)/(4-1) = -9/4

OpenStudy (unknownrandom):

\[\ln (x)=\frac{ -9 }{ 4 }\]

OpenStudy (unknownrandom):

c=\[e ^{-9/4}\] I thought that would be the right answer but it wasn't.

OpenStudy (unknownrandom):

-9/4 should be -1/4

OpenStudy (freckles):

why did you integrate 1/x and differentiate 1?

OpenStudy (unknownrandom):

How did I integrate?

OpenStudy (phi):

d/dx (x^-1) is not ln(x)

OpenStudy (freckles):

you can use power rule to differentiate 1/x aka x^(-1)

OpenStudy (freckles):

also I think your arithmetic was a little off when finding the average velocity of f on [a,b]

OpenStudy (unknownrandom):

Thanks guys. I guess I got them mixed up.

OpenStudy (freckles):

on [1,4]

OpenStudy (freckles):

\[\frac{\frac{5}{4}-2}{4-1} =\frac{5-8}{16-4}=?\] multiply top and bottom by 4 to make a bit easier on you

OpenStudy (unknownrandom):

-1/4 whoops...

OpenStudy (freckles):

so you should have the equation \[(1+\frac{1}{x})'=\frac{-1}{4} \\ \text{ where } (1+\frac{1}{x})' \neq \ln(x) \text{ of course }\]

OpenStudy (unknownrandom):

So, -1/x^2=-1/4

OpenStudy (freckles):

right or x^2=4 just remember you are looking for x in the interval [1,4]

OpenStudy (freckles):

or it really should be the interval (1,4)

OpenStudy (unknownrandom):

I got it. Thanks for the help @freckles! I was just making some simple mistakes.

OpenStudy (freckles):

np :) have a nice day

OpenStudy (unknownrandom):

@ freckles Can you help me solve this one when you get a chance? Farmer Alex has 32 llamas each of whom needs 1000000 square feet of grazing area. He wants to enclose a rectangular pen along a straight river, where he does not need fence. What is the minimum amount of fencing he must use to build his pen?

OpenStudy (unknownrandom):

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OpenStudy (unknownrandom):

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