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

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