concavity intervals and points of inflection?

Question

Hello I am in search of concavity intervals and points of inflection for F"(x)=30-12X/X-3. I thin kthere is only one point...2.5 but I could be totally wrong.

Thanks

Answer 1

Assuming that you are saying that F''(x) = (30-12x) / (x-3)

You find inflection points by finding where the graph changes from concave upward to concave downward (at F"(x) = 0 or where F"(x) does not exist).

0 = (30-12x) / (x-3)
0 = 30-12x
12x = 30
x = 2.5

**Inflection point at x = 2.5

F"(x) does not exist at x = 3, so there is an

**Inflection point at x = 3

The graph is concave upward on the interval where F"(x) > 0.
The graph is concave downward on the interval where F"(x) < 0.

So lets look at whether F"(x) is positive or negative on the intervals before, after, and between the two infelction points (2.5 and 3):

x < 2.5 :        F"(x) is negative, so concave downward on this interval
2.5 < x < 3 :   F"(x) is positive, so concave upward on this interval
x > 3 :          F"(x) is negative, so concave downward on this interval

The best way to approach these problems is to find all the x values that are possible inflection points (where f(x) =0 or does not exist). Then test an x-value in each interval between possible inflection points to see whether it produces a positive or negative value. If the value changes from positve to negative or negative to positive then we know the "possible" inflection point is for sure an inflection point.

Answer 2

The denominator obviously must be x -3, because it would make too little sense to have 12x/x. But what is the numerator?

Since lcamccandlj has made an assumption about the numerator, I will make the other one.

Assume you are saying F'(x) = 30 - 12/(x-3)

Setting that to 0 gives

30 = 12/(x-3)
30x - 90 = 12
30x = 102
x = 102/30 = 51/15

Answer 3

Points of inflection are indicated by the second derivate = 0. Concavity by the second derivative being negative. I cannot interpret your formula because of lack of brackets: Is it 30-12x/(x-3) or is it (30-12x)/(x-3). Is F*(x) the first derivative?

Answer 4

1. you know how to find horizontal limits and xandy intercepts etc right? because im just gonna skip into the derivative parts. y' = 2x+2 Set y' = 0 to find critical points. 2x+2 = 9 x= -1 Plug into f' along a number line to check whether we are increasing or decreasing. Decreasing (-∞,-1] Increasing [1.∞) <-- remember to be using interval nottation correctly. There is a local min at (-1, 6) No POI And concavity is up on interval (-∞,∞) Remember it is a parabolic function, you should be abel to handle number 2 now, dont forget to check your answers. just go onto wolframalpha and graph it

Answer 5

Just set the second derivative (which you have given) equal to zero. Thus,
F"(x) = 0

Then,
(30 - 12x)/(x - 3) = 0

We cross-multiply
30 - 12x = 0

Then,
-12x = -30

We divide,
x = -30/-12


Therefore,
x = 2.5

You are correct.

^_^

Answer 6

It would help if you add some parentheses to the equation for clarity.