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

- anonymous

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- anonymous

\[f(x) = \frac{ x^2+x }{ x-1 }\]

- anonymous

function is only discontinuous at 1, which is not in the interval
theorem holds

- 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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## More answers

- anonymous

\[f(x) = \frac{ x^2+x }{ x-1 }\] you need some numbers to get the second part

- anonymous

oh here

- anonymous

you need \(f(4)\) first

- anonymous

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

- anonymous

the statement \(f(c)=6\) makes no sense

- anonymous

well, it's given

- 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

oooh abort sorry wrong i am wrong

- anonymous

oh I didn't learn that yet but okay..
what should I do?

- anonymous

it thought it said mean value theorem, it says intermediate value theorem
we have to start over sorry

- anonymous

it's fine

- 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

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

since the function is continuous on that interval and cannot skip values

- anonymous

yes

- anonymous

sorry for the interruption but

- 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

you job is to solve
\[f(x)=6\] but that is for the second part of the question

- anonymous

there are two parts
1) verify that the intermediate value theorem applies to the indicated interval

- 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

i think \(f(\frac{5}{2})=5\tfrac{5}{6}\) and \(f(4)=\frac{20}{3}\)

- anonymous

how can 4 be larger than 6?

- anonymous

k lets go slow

- anonymous

it is not 4 that is larger than 6, it is \(f(4)=\frac{20}{3}\) that is larger than 6

- anonymous

in other words
\[f(\frac{5}{2})<6

- anonymous

okay

- 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

now solving it is just a matter of solving
\[\frac{x^2+x}{x-1}=6\]

- anonymous

you would start with
\[x^2+x=6(x-1)\] and solve the resulting quadratic equation

- anonymous

ooh
okay okay I finally get it :)

- anonymous

Thank you so much for your kind explanation! :)

- anonymous

yw

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