Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

oh here

Answer 6

you need \(f(4)\) first

Answer 7

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

Answer 8

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

Answer 9

well, it's given

Answer 10

mvt says under the suitable conditions
there is a \(c\in (a, b)\) where
\[f'(c)=\frac{f(b)-f(a)}{b-a}\]

Answer 11

oooh abort sorry wrong i am wrong

Answer 12

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

Answer 13

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

Answer 14

it's fine

Answer 15

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 \(f(4)\)

Answer 16

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

Answer 17

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

Answer 18

yes

Answer 19

sorry for the interruption but

Answer 20

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?

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

how can 4 be larger than 6?

Answer 26

k lets go slow

Answer 27

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

Answer 28

in other words
\[f(\frac{5}{2})<6<f(4)\]

Answer 29

okay

Answer 30

so since 6 lies between the two function values, there must be some
\[c\in (\frac{5}{2},4)\] where \(f(c)=6\)

Answer 31

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

Answer 32

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

Answer 33

ooh
okay okay I finally get it :)

Answer 34

Thank you so much for your kind explanation! :)

Answer 35

yw