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anonymous
 5 years ago
for: (10x3)/(2x+4)
I got 1st derivative as 34/(2x+4)^2
is that correct?
Next I need to find the Critical Number(s).
Would that be x=2?
anonymous
 5 years ago
for: (10x3)/(2x+4) I got 1st derivative as 34/(2x+4)^2 is that correct? Next I need to find the Critical Number(s). Would that be x=2?

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0your 1st derivative will never be zero, so it has no critical numbers..

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0it is undefined at x=2, but I would have to read up on that to see if it matter :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Amistre64, after you are done with this one, could you please take a look at the http://openstudy.com/groups/mathematics#/updates/4da5d0a9d6938b0b2e74a24d

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how do u go about it?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0check out your f(x) for starters; whats its domian?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0is it defined at x=2?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0if not, then what would the slope of a none existent point be?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0not f(x) = 2; but what is the value of f(2)? (10x3)/(2x+4) (10(2)3)/(2(2)+4) (203)/(4+4) 17/0 is f(2); which means that there is no value for f(2) to begin with. it has no slope since it is not a part of the graph, or the equation. f'(2) fails to exists because there is no point at f(2) in order to evaluate.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0is two the answer now?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I thought that the x=2 was not a critical number because it cannot be 0 in the den

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so if there is no critical number then to give the intervals of concavity, would I still test the x=2?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0there is no concavity at x=2 simply becasue x=2 has no value to the equation. If you have nothing to examine, then you simply can examine it :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0like this if I see it correctly

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes that is what my graph looks like. So on my work the question, I need to find Intervals of increase and decrease, so should I just say there are not any?

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the graph is increaseing on the interval (inf,2) the graph is decreasing on the interval (2,inf)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok on the graph you sent above are you using both of the lines. I don't see the increasing at (inf,2) and decreasing on interval (2,inf). Looking at it I would want to write (inf,2) increasing and (2,inf) increasing

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0hmm..... now that you mention it..... if we start at the left and move right, the slope of our line is increaseing as we mover from inf to 2. until at 2 it hits an unknown..... we hop over the line and the slope begins to decrease, gets less steep as we continue to move from 2 out to inf. So I was right, its increaseing on the left, and decreasing on the right.

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0the left is concave up, and the right is concave down.... but i havent really checked the 2nd derivative to prove it yet :)

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0one thing I notice tho is that the second derivative is always positive...which means the the slope is always in the positive condition, no matter how flat it gets, its always a positive slope.
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