## anonymous 3 years ago 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

1. anonymous

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

2. anonymous

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

3. 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

4. anonymous

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

5. anonymous

oh here

6. anonymous

you need $$f(4)$$ first

7. anonymous

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

8. anonymous

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

9. anonymous

well, it's given

10. anonymous

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

11. anonymous

oooh abort sorry wrong i am wrong

12. anonymous

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

13. anonymous

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

14. anonymous

it's fine

15. 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 $$f(4)$$

16. 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

17. anonymous

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

18. anonymous

yes

19. anonymous

sorry for the interruption but

20. 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?

21. anonymous

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

22. anonymous

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

23. 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

24. anonymous

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

25. anonymous

how can 4 be larger than 6?

26. anonymous

k lets go slow

27. anonymous

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

28. anonymous

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

29. anonymous

okay

30. anonymous

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

31. anonymous

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

32. anonymous

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

33. anonymous

ooh okay okay I finally get it :)

34. anonymous

Thank you so much for your kind explanation! :)

35. anonymous

yw