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 2 years 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]
 2 years 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]

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

mlddmlnog
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.1Understand how those got plugged in?

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

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

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

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

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

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

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

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

mlddmlnog
 2 years 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
 2 years 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
 2 years 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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