Question

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?

Answer 1

your 1st derivative will never be zero, so it has no critical numbers..

Answer 2

it is undefined at x=2, but I would have to read up on that to see if it matter :)

Answer 3

Amistre64, after you are done with this one, could you please take a look at the http://openstudy.com/groups/mathematics#/updates/4da5d0a9d6938b0b2e74a24d

Answer 4

how do u go about it?

Answer 5

check out your f(x) for starters; whats its domian?

Answer 6

is it defined at x=2?

Answer 7

if not, then what would the slope of a none existent point be?

Answer 8

how is f(x)=2?

Answer 9

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.

Answer 10

is two the answer now?

Answer 11

I thought that the x=2 was not a critical number because it cannot be 0 in the den

Answer 12

so if there is no critical number then to give the intervals of concavity, would I still test the x=2?

Answer 13

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 :)

Answer 14

like this if I see it correctly

Answer 15

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?

Answer 16

the graph is increaseing on the interval (-inf,2)

the graph is decreasing on the interval (2,inf)

Answer 17

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

Answer 18

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.

Answer 19

the left is concave up, and the right is concave down.... but i havent really checked the 2nd derivative to prove it yet :)

Answer 20

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.