anonymous
  • anonymous
for: (10x-3)/(-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?
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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amistre64
  • amistre64
your 1st derivative will never be zero, so it has no critical numbers..
amistre64
  • amistre64
it is undefined at x=2, but I would have to read up on that to see if it matter :)
anonymous
  • anonymous
Amistre64, after you are done with this one, could you please take a look at the http://openstudy.com/groups/mathematics#/updates/4da5d0a9d6938b0b2e74a24d

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anonymous
  • anonymous
how do u go about it?
amistre64
  • amistre64
check out your f(x) for starters; whats its domian?
amistre64
  • amistre64
is it defined at x=2?
amistre64
  • amistre64
if not, then what would the slope of a none existent point be?
anonymous
  • anonymous
how is f(x)=2?
amistre64
  • amistre64
not f(x) = 2; but what is the value of f(2)? (10x-3)/(-2x+4) (10(2)-3)/(-2(2)+4) (20-3)/(-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
  • anonymous
is two the answer now?
anonymous
  • anonymous
I thought that the x=2 was not a critical number because it cannot be 0 in the den
anonymous
  • anonymous
so if there is no critical number then to give the intervals of concavity, would I still test the x=2?
amistre64
  • amistre64
there 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
  • amistre64
like this if I see it correctly
1 Attachment
anonymous
  • anonymous
yes 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
  • amistre64
the graph is increaseing on the interval (-inf,2) the graph is decreasing on the interval (2,inf)
anonymous
  • anonymous
ok 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
  • amistre64
hmm..... 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
  • amistre64
the left is concave up, and the right is concave down.... but i havent really checked the 2nd derivative to prove it yet :)
amistre64
  • amistre64
one 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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