At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
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?
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
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.
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?
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 solved it...well after you help of course...i started to see the problem working out...i got another...
I was gonna square both sides and do some magic :)
\[f(x)=(x-3)(x-7)^3+8\] we are looking for absolute max and min for intervals at [1,4] and [1,8] and [4,9]
.... i got 9/2...... so nyahh!!
.......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???
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......
at 5 and 7 we have inflection points...
ok but our first interval were looking at is [1,4]
<..................5......................7....................> --- +++ +++ --- --- +++ ---------------------------------- + - + when we multiply these "zones" together, we get these results
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?
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
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...
start a new question, I think this ones lagging alot becasue of all the "replies" its trying to keep track of :)