Question

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.

Answer 1

this one here?

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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.

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

ok then what next?

Answer 11

whats our mean slope? I get:

4sqrt(2)
------
3

Answer 12

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

Answer 13

thats what i got...it was correct

Answer 14

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

Answer 15

4/sqrt(x) maybe?

Answer 16

yes it is

Answer 17

yay!!...

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

Answer 18

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

3 sqrt(x) = x sqrt(2)

Answer 19

the answer is 4.5....

Answer 20

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

Answer 21

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

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

same function?

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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

Answer 30

f(x) = x^4 -24x^3 +210x^2 -784x +1037

f'(x) = 4x^3 -72x^2 +420x -784

:)

Answer 31

your way might be easier to read :)

Answer 32

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

Answer 33

yep..

Answer 34

x = 7 is a zero, thats easy to see

Answer 35

and would 3 be a zero

Answer 36

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

Answer 37

take the second derivative to see how the graph is behaving

Answer 38

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

Answer 39

factor out a 12..

12(x^2 - 12x + 35)

Answer 40

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

Answer 41

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

Answer 42

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

Answer 43

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

Answer 44

at 5 and 7 we have inflection points...

Answer 45

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

Answer 46

<..................5......................7....................>
--- +++ +++
--- --- +++
----------------------------------
+ - +

when we multiply these "zones" together, we get these results

Answer 47

do you undertand this diagram?

Answer 48

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

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

our graph is acting like this:

Answer 53

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

Answer 54

f(1) = 440 if I did it right

Answer 55

f(4) = -19

Answer 56

the max is at (1,440)

Answer 57

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

Answer 58

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

Answer 59

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

Answer 60

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

Answer 61

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

Answer 62

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

Answer 63

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