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this one here?

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

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

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

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

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

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

ok then what next?

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

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

thats what i got...it was correct

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

4/sqrt(x) maybe?

yes it is

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

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

the answer is 4.5....

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

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

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

same function?

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

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

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

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

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

your way might be easier to read :)

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

yep..

x = 7 is a zero, thats easy to see

and would 3 be a zero

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

take the second derivative to see how the graph is behaving

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

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

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

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

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

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

at 5 and 7 we have inflection points...

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

do you undertand this diagram?

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

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

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

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

f(1) = 440 if I did it right

f(4) = -19

the max is at (1,440)

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

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

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

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

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

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