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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)/(x-1), [5/2,4], f(c) = 6

Mathematics
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\[f(x) = \frac{ x^2+x }{ x-1 }\]
function is only discontinuous at 1, which is not in the interval theorem holds
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

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\[f(x) = \frac{ x^2+x }{ x-1 }\] you need some numbers to get the second part
oh here
you need \(f(4)\) first
[5/2 , 4] f(c) = 6
the statement \(f(c)=6\) makes no sense
well, it's given
mvt says under the suitable conditions there is a \(c\in (a, b)\) where \[f'(c)=\frac{f(b)-f(a)}{b-a}\]
oooh abort sorry wrong i am wrong
oh I didn't learn that yet but okay.. what should I do?
it thought it said mean value theorem, it says intermediate value theorem we have to start over sorry
it's fine
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 \(f(4)\)
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
since the function is continuous on that interval and cannot skip values
yes
sorry for the interruption but
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?
you job is to solve \[f(x)=6\] but that is for the second part of the question
there are two parts 1) verify that the intermediate value theorem applies to the indicated interval
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
i think \(f(\frac{5}{2})=5\tfrac{5}{6}\) and \(f(4)=\frac{20}{3}\)
how can 4 be larger than 6?
k lets go slow
it is not 4 that is larger than 6, it is \(f(4)=\frac{20}{3}\) that is larger than 6
in other words \[f(\frac{5}{2})<6
okay
so since 6 lies between the two function values, there must be some \[c\in (\frac{5}{2},4)\] where \(f(c)=6\)
now solving it is just a matter of solving \[\frac{x^2+x}{x-1}=6\]
you would start with \[x^2+x=6(x-1)\] and solve the resulting quadratic equation
ooh okay okay I finally get it :)
Thank you so much for your kind explanation! :)
yw

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