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## anonymous 5 years ago Consider the function f(x) = 8 √{ x} + 9 on the interval [ 2 , 8 ] (A) Find the average or mean slope of the function on this interval (B) By the Mean Value Theorem, we know there exists at least one c in the open interval ( 2 , 8 ) such that f′( c) is equal to this mean slope. Find all values of c that work and list them (separated by commas) in the box below.

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1. amistre64

this one here?

2. anonymous

yea this is one of the ones i am having trouble on...

3. amistre64

mean slope means what? the slope of the line from end point to end point?

4. anonymous

i don't know i check all my notes and no where do i see mean slope so i don't know why the teacher assigned this problem

5. amistre64

I believe that it means the average slope between the intervals and is found by taking the slope of the line between the end points.

6. amistre64

then you find a point in the interval that touches the graph at one point with that slope

7. anonymous

so what would be the equation to find the slope between the intervals

8. amistre64

f(8) - f(2) change in y -------- over 8-2 change in x

9. amistre64

this is how you find any slope of a line :) the change in y over the change in x

10. anonymous

ok then what next?

11. amistre64

whats our mean slope? I get: 4sqrt(2) ------ 3

12. amistre64

when f'(c) = 4sqrt(2)/3 then we have found the values for "x" that satisfy the problem

13. anonymous

thats what i got...it was correct

14. amistre64

f(x) = 8x^(1/2) + 9 whats our derivative?

15. amistre64

4/sqrt(x) maybe?

16. anonymous

yes it is

17. amistre64

yay!!... 4sqrt(x) 4sqrt(2) ------ = -------- we need to solve this then x 3

18. amistre64

12 sqrt(x) = 4x sqrt(2) ; divide both sides by 4 3 sqrt(x) = x sqrt(2)

19. anonymous

the answer is 4.5....

20. amistre64

did you solve it, or is that what we are going for?

21. anonymous

i solved it...well after you help of course...i started to see the problem working out...i got another...

22. amistre64

I was gonna square both sides and do some magic :)

23. anonymous

$f(x)=(x-3)(x-7)^3+8$ we are looking for absolute max and min for intervals at [1,4] and [1,8] and [4,9]

24. amistre64

.... i got 9/2...... so nyahh!!

25. amistre64

same function?

26. amistre64

.......ok, i can read..... sometimes :)

27. amistre64

its a quartic equation which will resemble a "W" perhaps

28. amistre64

just expand it and make it easier for ya...

29. anonymous

ok..lol... i got $f'(x)=(x-7)^3+(x-3)(3)(x-7)^2$

30. amistre64

f(x) = x^4 -24x^3 +210x^2 -784x +1037 f'(x) = 4x^3 -72x^2 +420x -784 :)

31. amistre64

your way might be easier to read :)

32. anonymous

ok so i would make the equation equal to zero right???

33. amistre64

yep..

34. amistre64

x = 7 is a zero, thats easy to see

35. anonymous

and would 3 be a zero

36. amistre64

3 would be a zero + (-7)^3

37. amistre64

take the second derivative to see how the graph is behaving

38. amistre64

f''(x) = 12x^2 -144x +420

39. amistre64

factor out a 12.. 12(x^2 - 12x + 35)

40. amistre64

12(x-7)(x-5) is the factored form

41. anonymous

ok and how would i use that to find the max and min between all three intervals

42. amistre64

draw a number line straight across the paper......

43. amistre64

<..................5......................7....................>

44. amistre64

at 5 and 7 we have inflection points...

45. anonymous

ok but our first interval were looking at is [1,4]

46. amistre64

<..................5......................7....................> --- +++ +++ --- --- +++ ---------------------------------- + - + when we multiply these "zones" together, we get these results

47. amistre64

do you undertand this diagram?

48. anonymous

yea...how do we apply the intervals to this...

49. amistre64

[1,4] is in an area that is concave up, so if anything it would have a MIN in it someplace

50. amistre64

Now, are we looking for the "highest and or lowest point in the intervals?

51. anonymous

i am going to break up the question like it is.... f(x)=(x−3)(x−7)^3+8 a. interval [1,4] absolute max: absolute min: this is how all of the other ones are

52. amistre64

our graph is acting like this:

53. amistre64

solve for f(1) and f(4) to see what our values are.

54. amistre64

f(1) = 440 if I did it right

55. amistre64

f(4) = -19

56. amistre64

the max is at (1,440)

57. amistre64

here a thought, plug in 4 into the derivative and see if we get a zero :)

58. amistre64

yep, 4 is a zero for f'(x)...im a genuis :)

59. amistre64

[1,4] (1 ,440) = max and (4,-19) = min you agree?

60. amistre64

and the graph doesnt have an more "bumps in it, just some transitional bends.....

61. anonymous

yea the next interval is the same and the last one the max is (9,56) and min is (4,-19)

62. anonymous

$f(x)=(3+9\ln (x))/x$ i need to find the x coordinate of the absolute max...

63. amistre64

start a new question, I think this ones lagging alot becasue of all the "replies" its trying to keep track of :)

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