## 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 11 months ago 11 months ago

1. musiklover317 Group Title

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

2. satellite73 Group Title

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

3. musiklover317 Group Title

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 Group Title

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

5. musiklover317 Group Title

oh here

6. satellite73 Group Title

you need $$f(4)$$ first

7. musiklover317 Group Title

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

8. satellite73 Group Title

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

9. musiklover317 Group Title

well, it's given

10. satellite73 Group Title

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 Group Title

oooh abort sorry wrong i am wrong

12. musiklover317 Group Title

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

13. satellite73 Group Title

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

14. musiklover317 Group Title

it's fine

15. satellite73 Group Title

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 Group Title

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 Group Title

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

18. musiklover317 Group Title

yes

19. musiklover317 Group Title

sorry for the interruption but

20. musiklover317 Group Title

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 Group Title

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

22. satellite73 Group Title

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

23. satellite73 Group Title

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 Group Title

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

25. musiklover317 Group Title

how can 4 be larger than 6?

26. satellite73 Group Title

k lets go slow

27. satellite73 Group Title

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

28. satellite73 Group Title

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

29. musiklover317 Group Title

okay

30. satellite73 Group Title

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 Group Title

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

32. satellite73 Group Title

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

33. musiklover317 Group Title

ooh okay okay I finally get it :)

34. musiklover317 Group Title

Thank you so much for your kind explanation! :)

35. satellite73 Group Title

yw