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anonymous
 3 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
anonymous
 3 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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anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[f(x) = \frac{ x^2+x }{ x1 }\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0function is only discontinuous at 1, which is not in the interval theorem holds

anonymous
 3 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\[f(x) = \frac{ x^2+x }{ x1 }\] you need some numbers to get the second part

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you need \(f(4)\) first

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the statement \(f(c)=6\) makes no sense

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0mvt says under the suitable conditions there is a \(c\in (a, b)\) where \[f'(c)=\frac{f(b)f(a)}{ba}\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0oooh abort sorry wrong i am wrong

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it thought it said mean value theorem, it says intermediate value theorem we have to start over sorry

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0they 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 \(f(4)\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0one 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0since the function is continuous on that interval and cannot skip values

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

anonymous
 3 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?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you job is to solve \[f(x)=6\] but that is for the second part of the question

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0there are two parts 1) verify that the intermediate value theorem applies to the indicated interval

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i think \(f(\frac{5}{2})=5\tfrac{5}{6}\) and \(f(4)=\frac{20}{3}\)

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0it is not 4 that is larger than 6, it is \(f(4)=\frac{20}{3}\) that is larger than 6

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0in other words \[f(\frac{5}{2})<6<f(4)\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so since 6 lies between the two function values, there must be some \[c\in (\frac{5}{2},4)\] where \(f(c)=6\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now solving it is just a matter of solving \[\frac{x^2+x}{x1}=6\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0you would start with \[x^2+x=6(x1)\] and solve the resulting quadratic equation

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

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