## musiklover317 Group Title 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 one year ago one year ago

1. musiklover317

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

2. satellite73

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

3. musiklover317

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

4. satellite73

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

5. musiklover317

oh here

6. satellite73

you need $$f(4)$$ first

7. musiklover317

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

8. satellite73

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

9. musiklover317

well, it's given

10. satellite73

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

11. satellite73

oooh abort sorry wrong i am wrong

12. musiklover317

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

13. satellite73

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

14. musiklover317

it's fine

15. satellite73

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 $$f(4)$$

16. satellite73

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

17. satellite73

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

18. musiklover317

yes

19. musiklover317

sorry for the interruption but

20. musiklover317

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?

21. satellite73

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

22. satellite73

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

23. satellite73

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

24. satellite73

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

25. musiklover317

how can 4 be larger than 6?

26. satellite73

k lets go slow

27. satellite73

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

28. satellite73

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

29. musiklover317

okay

30. satellite73

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

31. satellite73

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

32. satellite73

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

33. musiklover317

ooh okay okay I finally get it :)

34. musiklover317

Thank you so much for your kind explanation! :)

35. satellite73

yw