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mlddmlnog

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
  • one year ago

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  1. mlddmlnog
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    @zepdrix

    • one year ago
  2. zepdrix
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    Do you understand how the Meal-Value Theorem works? :O

    • one year ago
  3. mlddmlnog
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    hm.. i'll try to do this one myself and ask for help if i still can't get the answer :)

    • one year ago
  4. zepdrix
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    |dw:1358281444723:dw|Ok but here's a brief explanation of the concept c: just in case it helps.

    • one year ago
  5. zepdrix
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    |dw:1358281503651:dw|

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

    • one year ago
  7. mlddmlnog
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    ok, i need help... i dont know how to do this.. haha i thought i did but i dont. so, please continueee :)

    • one year ago
  8. zepdrix
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    So 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.\[\large m=\frac{f(6)-f(-4)}{6-(-4)}\]

    • one year ago
  9. zepdrix
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    \[\large m=\frac{\color{orangered}{f(6)}-\color{cornflowerblue}{f(-4)}}{6-(-4)}\] \[\large m=\frac{(\color{orangered}{6^2+6})-(\color{cornflowerblue}{(-4)^2-4})}{6+4}\]

    • one year ago
  10. zepdrix
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    Understand how those got plugged in?

    • one year ago
  11. mlddmlnog
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    YES :)

    • one year ago
  12. zepdrix
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    Simplifying that down, I think we end up withhhhhhhhh 3ish.

    • one year ago
  13. mlddmlnog
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    yes. i got 3.

    • one year ago
  14. zepdrix
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    So 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.

    • one year ago
  15. zepdrix
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    Let's first find f'(x).

    • one year ago
  16. zepdrix
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    What'd u get? :O

    • one year ago
  17. mlddmlnog
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    2x+1 :)

    • one year ago
  18. zepdrix
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    k cool :D

    • one year ago
  19. zepdrix
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    So then,\[\large f'(c)=2c+1\]And we're claiming, that by the Mean-Value 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. \[\large 2c+1=3\]

    • one year ago
  20. mlddmlnog
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    c=1

    • one year ago
  21. zepdrix
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    When 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.

    • one year ago
  22. mlddmlnog
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    since it's 1, it is between -4 and 6 :)

    • one year ago
  23. zepdrix
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    Yay team c:

    • one year ago
  24. mlddmlnog
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    yes. 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 anti-derivative of F on [a,b], then \[\int\limits_{a}^{b}f(x)dx=f(b)-f(a)\].

    • one year ago
  25. mlddmlnog
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    oops, i meant [a,b]..

    • one year ago
  26. zepdrix
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    Oh crap maybe i confused it with the intermediate value theorem... sec.. thinking :3

    • one year ago
  27. mlddmlnog
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    ahh darn. haha ok ;p

    • one year ago
  28. zepdrix
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    hmm no it doesn't appear i did... hmmmm

    • one year ago
  29. zepdrix
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    The 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

    • one year ago
  30. mlddmlnog
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    oh 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 \[\int\limits_{a}^{b}f(x)dx=f(x*)(b-a).\]

    • one year ago
  31. mlddmlnog
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    if f is continuous on [a,b] i meant.

    • one year ago
  32. zepdrix
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    Hmm crap I dunno D: I'm familiar with seeing it that way.

    • one year ago
  33. mlddmlnog
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    aw darn it.. is there anyone you could ask help for? because that formula is what my teacher uses... :'(

    • one year ago
  34. zepdrix
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    @hartnn @hba @campbell_st Maybe one of these fellas know what you're talking about c:

    • one year ago
  35. mlddmlnog
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    ahh thankss :)

    • one year ago
  36. mlddmlnog
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    oh nvermind! we got the correct answer. so i';lljust do it the way @zepdrix did it! :D thank you for comming anyway! :)

    • one year ago
  37. hartnn
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    ohh..okay :)

    • one year ago
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