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musiklover317
Group Title
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
 10 months ago
 10 months ago
musiklover317 Group Title
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
 10 months ago
 10 months ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
function is only discontinuous at 1, which is not in the interval theorem holds
 10 months ago

musiklover317 Group TitleBest ResponseYou've already chosen the best response.0
hmm... 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
 10 months ago

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

musiklover317 Group TitleBest ResponseYou've already chosen the best response.0
oh here
 10 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
you need \(f(4)\) first
 10 months ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
the statement \(f(c)=6\) makes no sense
 10 months ago

musiklover317 Group TitleBest ResponseYou've already chosen the best response.0
well, it's given
 10 months ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
oooh abort sorry wrong i am wrong
 10 months ago

musiklover317 Group TitleBest ResponseYou've already chosen the best response.0
oh I didn't learn that yet but okay.. what should I do?
 10 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
it thought it said mean value theorem, it says intermediate value theorem we have to start over sorry
 10 months ago

musiklover317 Group TitleBest ResponseYou've already chosen the best response.0
it's fine
 10 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
they 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 \(f(4)\)
 10 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
one 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
 10 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
since the function is continuous on that interval and cannot skip values
 10 months ago

musiklover317 Group TitleBest ResponseYou've already chosen the best response.0
yes
 10 months ago

musiklover317 Group TitleBest ResponseYou've already chosen the best response.0
sorry for the interruption but
 10 months ago

musiklover317 Group TitleBest ResponseYou've already chosen the best response.0
well, 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?
 10 months ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
there are two parts 1) verify that the intermediate value theorem applies to the indicated interval
 10 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
the 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
 10 months ago

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

musiklover317 Group TitleBest ResponseYou've already chosen the best response.0
how can 4 be larger than 6?
 10 months ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
k lets go slow
 10 months ago

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

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

musiklover317 Group TitleBest ResponseYou've already chosen the best response.0
okay
 10 months ago

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

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

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

musiklover317 Group TitleBest ResponseYou've already chosen the best response.0
ooh okay okay I finally get it :)
 10 months ago

musiklover317 Group TitleBest ResponseYou've already chosen the best response.0
Thank you so much for your kind explanation! :)
 10 months ago
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