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 one year ago
verify that the hypothesis of the Mean value theorem are satisfied on the given interval and findall values of c that satisfy theconclusion of the theorem.
f(x)=x^2+x [4,6]
 one year ago
verify that the hypothesis of the Mean value theorem are satisfied on the given interval and findall values of c that satisfy theconclusion of the theorem. f(x)=x^2+x [4,6]

This Question is Closed

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Do you understand how the MealValue Theorem works? :O

mlddmlnog
 one year ago
Best ResponseYou've already chosen the best response.0hm.. i'll try to do this one myself and ask for help if i still can't get the answer :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1dw:1358281444723:dwOk but here's a brief explanation of the concept c: just in case it helps.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Remember the formula for finding the slope of a line between 2 points? (a secant line) If we use the notation from the graph,\[\large m=\frac{f(b)f(a)}{ba}\] In order for the function to be continuous, this slope must be equal to the slope of a TANGENT line somewhere in the middle.\[\large \frac{f(b)f(a)}{ba}=f'(c)\] A derivative at a particular point represents the slope of a tangent line.

mlddmlnog
 one year ago
Best ResponseYou've already chosen the best response.0ok, i need help... i dont know how to do this.. haha i thought i did but i dont. so, please continueee :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So in this problem we're starting with a secant line, passing through x=4 and x=6. Let's start by finding the slope \(m\) of that line.\[\large m=\frac{f(6)f(4)}{6(4)}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large m=\frac{\color{orangered}{f(6)}\color{cornflowerblue}{f(4)}}{6(4)}\] \[\large m=\frac{(\color{orangered}{6^2+6})(\color{cornflowerblue}{(4)^24})}{6+4}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Understand how those got plugged in?

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Simplifying that down, I think we end up withhhhhhhhh 3ish.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So the theorem says  In order for this function to be continuous from 4 to 6, there must be a TANGENT line somewhere between those points that has this slope of 3.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Let's first find f'(x).

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So then,\[\large f'(c)=2c+1\]And we're claiming, that by the MeanValue Theorem this tangent line at x=c has a slope of 3. It has to happen somewhere between 4 and 6 in order for the function to be continuous. So let's set our f'(c) equal to 3 and solve for c. \[\large 2c+1=3\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1When you get an answer, verify whether or not it is BETWEEN 4 and 6. Because that will help to verify that we didn't make a mistake somewhere.

mlddmlnog
 one year ago
Best ResponseYou've already chosen the best response.0since it's 1, it is between 4 and 6 :)

mlddmlnog
 one year ago
Best ResponseYou've already chosen the best response.0yes. but is that the final answer? i thought we were supposed to use the mean value theorem.. which says: if F is continuous on [z,b], and if F is an antiderivative of F on [a,b], then \[\int\limits_{a}^{b}f(x)dx=f(b)f(a)\].

mlddmlnog
 one year ago
Best ResponseYou've already chosen the best response.0oops, i meant [a,b]..

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Oh crap maybe i confused it with the intermediate value theorem... sec.. thinking :3

mlddmlnog
 one year ago
Best ResponseYou've already chosen the best response.0ahh darn. haha ok ;p

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1hmm no it doesn't appear i did... hmmmm

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1The thing you posted is the FTC (Fundamental Theorem of Calculus, Part 2).\[\large \int\limits_a^b f(x)dx=F(b)F(a)\] I'm not quite sure what that has to do with the mean value theorem D: hmmm

mlddmlnog
 one year ago
Best ResponseYou've already chosen the best response.0oh sorryyy yea thta IS the fundamental theorem of calculus.. ha but the mean value theorem of integrals says : if f is conb], then there exists at least one number x* in [a,b] such that \[\int\limits_{a}^{b}f(x)dx=f(x*)(ba).\]

mlddmlnog
 one year ago
Best ResponseYou've already chosen the best response.0if f is continuous on [a,b] i meant.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Hmm crap I dunno D: I'm familiar with seeing it that way.

mlddmlnog
 one year ago
Best ResponseYou've already chosen the best response.0aw darn it.. is there anyone you could ask help for? because that formula is what my teacher uses... :'(

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1@hartnn @hba @campbell_st Maybe one of these fellas know what you're talking about c:

mlddmlnog
 one year ago
Best ResponseYou've already chosen the best response.0oh nvermind! we got the correct answer. so i';lljust do it the way @zepdrix did it! :D thank you for comming anyway! :)
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