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.

- anonymous

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- amistre64

this one here?

- anonymous

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

- amistre64

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

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## More answers

- 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

- 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.

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- anonymous

ok then what next?

- amistre64

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

- amistre64

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

- anonymous

thats what i got...it was correct

- amistre64

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

- amistre64

4/sqrt(x) maybe?

- anonymous

yes it is

- amistre64

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

- amistre64

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

- anonymous

the answer is 4.5....

- amistre64

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

- anonymous

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

- amistre64

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

- 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]

- amistre64

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

- amistre64

same function?

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

your way might be easier to read :)

- anonymous

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

- amistre64

yep..

- amistre64

x = 7 is a zero, thats easy to see

- anonymous

and would 3 be a zero

- amistre64

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

- amistre64

take the second derivative to see how the graph is behaving

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

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

- amistre64

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

- amistre64

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

- amistre64

at 5 and 7 we have inflection points...

- anonymous

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

- amistre64

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

- amistre64

do you undertand this diagram?

- anonymous

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

- amistre64

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

- amistre64

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

- 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

- amistre64

our graph is acting like this:

##### 1 Attachment

- amistre64

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

- amistre64

f(1) = 440 if I did it right

- amistre64

f(4) = -19

- amistre64

the max is at (1,440)

- amistre64

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

- amistre64

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

- amistre64

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

- amistre64

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

- anonymous

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

- anonymous

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

- 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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