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your 1st derivative will never be zero, so it has no critical numbers..
it is undefined at x=2, but I would have to read up on that to see if it matter :)
how do u go about it?
check out your f(x) for starters; whats its domian?
is it defined at x=2?
if not, then what would the slope of a none existent point be?
how is f(x)=2?
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.
is two the answer now?
I thought that the x=2 was not a critical number because it cannot be 0 in the den
so if there is no critical number then to give the intervals of concavity, would I still test the x=2?
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 :)
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?
the graph is increaseing on the interval (-inf,2) the graph is decreasing on the interval (2,inf)
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
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.
the left is concave up, and the right is concave down.... but i havent really checked the 2nd derivative to prove it yet :)
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.