anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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amistre64
  • amistre64
this one here?
anonymous
  • anonymous
yea this is one of the ones i am having trouble on...
amistre64
  • amistre64
mean slope means what? the slope of the line from end point to end point?

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More answers

anonymous
  • 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
  • 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
  • amistre64
then you find a point in the interval that touches the graph at one point with that slope
anonymous
  • anonymous
so what would be the equation to find the slope between the intervals
amistre64
  • amistre64
f(8) - f(2) change in y -------- over 8-2 change in x
amistre64
  • amistre64
this is how you find any slope of a line :) the change in y over the change in x
anonymous
  • anonymous
ok then what next?
amistre64
  • amistre64
whats our mean slope? I get: 4sqrt(2) ------ 3
amistre64
  • amistre64
when f'(c) = 4sqrt(2)/3 then we have found the values for "x" that satisfy the problem
anonymous
  • anonymous
thats what i got...it was correct
amistre64
  • amistre64
f(x) = 8x^(1/2) + 9 whats our derivative?
amistre64
  • amistre64
4/sqrt(x) maybe?
anonymous
  • anonymous
yes it is
amistre64
  • amistre64
yay!!... 4sqrt(x) 4sqrt(2) ------ = -------- we need to solve this then x 3
amistre64
  • amistre64
12 sqrt(x) = 4x sqrt(2) ; divide both sides by 4 3 sqrt(x) = x sqrt(2)
anonymous
  • anonymous
the answer is 4.5....
amistre64
  • amistre64
did you solve it, or is that what we are going for?
anonymous
  • anonymous
i solved it...well after you help of course...i started to see the problem working out...i got another...
amistre64
  • amistre64
I was gonna square both sides and do some magic :)
anonymous
  • 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
  • amistre64
.... i got 9/2...... so nyahh!!
amistre64
  • amistre64
same function?
amistre64
  • amistre64
.......ok, i can read..... sometimes :)
amistre64
  • amistre64
its a quartic equation which will resemble a "W" perhaps
amistre64
  • amistre64
just expand it and make it easier for ya...
anonymous
  • anonymous
ok..lol... i got \[f'(x)=(x-7)^3+(x-3)(3)(x-7)^2\]
amistre64
  • amistre64
f(x) = x^4 -24x^3 +210x^2 -784x +1037 f'(x) = 4x^3 -72x^2 +420x -784 :)
amistre64
  • amistre64
your way might be easier to read :)
anonymous
  • anonymous
ok so i would make the equation equal to zero right???
amistre64
  • amistre64
yep..
amistre64
  • amistre64
x = 7 is a zero, thats easy to see
anonymous
  • anonymous
and would 3 be a zero
amistre64
  • amistre64
3 would be a zero + (-7)^3
amistre64
  • amistre64
take the second derivative to see how the graph is behaving
amistre64
  • amistre64
f''(x) = 12x^2 -144x +420
amistre64
  • amistre64
factor out a 12.. 12(x^2 - 12x + 35)
amistre64
  • amistre64
12(x-7)(x-5) is the factored form
anonymous
  • anonymous
ok and how would i use that to find the max and min between all three intervals
amistre64
  • amistre64
draw a number line straight across the paper......
amistre64
  • amistre64
<..................5......................7....................>
amistre64
  • amistre64
at 5 and 7 we have inflection points...
anonymous
  • anonymous
ok but our first interval were looking at is [1,4]
amistre64
  • amistre64
<..................5......................7....................> --- +++ +++ --- --- +++ ---------------------------------- + - + when we multiply these "zones" together, we get these results
amistre64
  • amistre64
do you undertand this diagram?
anonymous
  • anonymous
yea...how do we apply the intervals to this...
amistre64
  • amistre64
[1,4] is in an area that is concave up, so if anything it would have a MIN in it someplace
amistre64
  • amistre64
Now, are we looking for the "highest and or lowest point in the intervals?
anonymous
  • 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
  • amistre64
our graph is acting like this:
1 Attachment
amistre64
  • amistre64
solve for f(1) and f(4) to see what our values are.
amistre64
  • amistre64
f(1) = 440 if I did it right
amistre64
  • amistre64
f(4) = -19
amistre64
  • amistre64
the max is at (1,440)
amistre64
  • amistre64
here a thought, plug in 4 into the derivative and see if we get a zero :)
amistre64
  • amistre64
yep, 4 is a zero for f'(x)...im a genuis :)
amistre64
  • amistre64
[1,4] (1 ,440) = max and (4,-19) = min you agree?
amistre64
  • amistre64
and the graph doesnt have an more "bumps in it, just some transitional bends.....
anonymous
  • anonymous
yea the next interval is the same and the last one the max is (9,56) and min is (4,-19)
anonymous
  • anonymous
\[f(x)=(3+9\ln (x))/x\] i need to find the x coordinate of the absolute max...
amistre64
  • 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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