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musiklover317
 2 years ago
pleasee help
verify that the intermediate value theorem applies to the indicated interval and find the value of c guaranteed by the theorem.
f(x) = (x^2+x)/(x1), [5/2,4], f(c) = 6
musiklover317
 2 years ago
pleasee help verify that the intermediate value theorem applies to the indicated interval and find the value of c guaranteed by the theorem. f(x) = (x^2+x)/(x1), [5/2,4], f(c) = 6

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musiklover317
 2 years ago
Best ResponseYou've already chosen the best response.0\[f(x) = \frac{ x^2+x }{ x1 }\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1function is only discontinuous at 1, which is not in the interval theorem holds

musiklover317
 2 years ago
Best ResponseYou've already chosen the best response.0hmm... for the other questions, I plugged in c to x and set that number (in this case, 6) equal to the equation and solved it. but for this one, I can't

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1\[f(x) = \frac{ x^2+x }{ x1 }\] you need some numbers to get the second part

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1you need \(f(4)\) first

musiklover317
 2 years ago
Best ResponseYou've already chosen the best response.0[5/2 , 4] f(c) = 6

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1the statement \(f(c)=6\) makes no sense

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1mvt says under the suitable conditions there is a \(c\in (a, b)\) where \[f'(c)=\frac{f(b)f(a)}{ba}\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1oooh abort sorry wrong i am wrong

musiklover317
 2 years ago
Best ResponseYou've already chosen the best response.0oh I didn't learn that yet but okay.. what should I do?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1it thought it said mean value theorem, it says intermediate value theorem we have to start over sorry

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1they way you show there must be a number in \([\frac{5}{2},4]\) with \[f(c)=6\] is to compute \(f(\frac{5}{2})\) and \(f(4)\)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1one number will be larger than 6, one number will be smaller than 6, so by the intermediate value theorem the function must be 6 somewhere in that interval

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1since the function is continuous on that interval and cannot skip values

musiklover317
 2 years ago
Best ResponseYou've already chosen the best response.0sorry for the interruption but

musiklover317
 2 years ago
Best ResponseYou've already chosen the best response.0well, in class, I learned that c has to be replaced with x and have to set that equation to 6. when I get the x values, the values between 5/2 and 4 is the answer is this right?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1you job is to solve \[f(x)=6\] but that is for the second part of the question

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1there are two parts 1) verify that the intermediate value theorem applies to the indicated interval

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1the reason that you know \[f(x)=6\] for some \(x\) in the interval is because \(f(4)\) is larger than 6, and \(f(\frac{5}{2})\) is smaller than 6

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1i think \(f(\frac{5}{2})=5\tfrac{5}{6}\) and \(f(4)=\frac{20}{3}\)

musiklover317
 2 years ago
Best ResponseYou've already chosen the best response.0how can 4 be larger than 6?

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1it is not 4 that is larger than 6, it is \(f(4)=\frac{20}{3}\) that is larger than 6

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1in other words \[f(\frac{5}{2})<6<f(4)\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1so since 6 lies between the two function values, there must be some \[c\in (\frac{5}{2},4)\] where \(f(c)=6\)

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1now solving it is just a matter of solving \[\frac{x^2+x}{x1}=6\]

satellite73
 2 years ago
Best ResponseYou've already chosen the best response.1you would start with \[x^2+x=6(x1)\] and solve the resulting quadratic equation

musiklover317
 2 years ago
Best ResponseYou've already chosen the best response.0ooh okay okay I finally get it :)

musiklover317
 2 years ago
Best ResponseYou've already chosen the best response.0Thank you so much for your kind explanation! :)
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