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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

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

oooh abort sorry wrong i am wrong

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

it's fine

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

yes

sorry for the interruption but

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

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

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