Question

(x+5)(2x-4)^-1 intervals where the function is increasing or decreasin, inflection points and concavity

Answer 1

did you take the derivative?

Answer 2

using quotient rule you will get
\[-\frac{7}{(2x-4)^2}\]

Answer 3

and a little thought will convince you that this thing is always negative, since the denominator is a perfect square.

Answer 4

therefore the original function is always decreasing. except of course at 2 where it is undefined

Answer 5

then take the derivative again

Answer 6

so it does not have a local maximum or minimum?

Answer 7

this time you get
\[\frac{7}{(2x-4)^3}\]

Answer 8

nope it is always decreasing. try graphing it and you will see

Answer 9

has vertical asymptote at x = 2, horizontal asymptote at y = 1/2

Answer 10

you can graph here and see it
http://www.wolframalpha.com/input/?i=y%3D%28x%2B5%29%2F%282x-4%29

Answer 11

always decreasing. we know this because derivative is always negative. so no local max or min

Answer 12

you can also see from the picture that it is "concave down" on x < 2 and concave up on x > 2

Answer 13

that is because
\[\frac{7}{(2x-4)^3}\] is negative if x < 2 and positive if x > 2

Answer 14

and the inflection point would be at x=2?

Answer 15

yes

Answer 16

I don't actually know, there is a discontinuity there, is it still an inflection point?

Answer 17

as far as i recall "inflection point" means where function changes concavity.

Answer 18

I think perhaps not, 2 is not in the domain.

Answer 19

so doesn't need to be defined there. just where it goes from being concave up to concave down or vice versa

Answer 20

But that is 2, which is not in the domain.

Answer 21

but i will concede to being wrong for sure

Answer 22

Normally you want to set the second derivative to 0, a bit difficult here.

Answer 23

so maybe it is incorrect to say that 2 is an inflection point. maybe it is simply that f is concave down over one interval and concave up over another. i could easily be incorrect.

Answer 24

ok now i know i am incorrect.
http://en.wikipedia.org/wiki/Inflection_point

Answer 25

Where does it say?

Answer 26

Ok, I see it now.

Answer 27

f''(x) should equall zero to have a inflection point?

Answer 28

Conclusion: function is asymptotic at 2 and there is no point of inflection even though the concavity changes before and after.

Answer 29

i think the point is that the function should be defined there. it is not necessary for
\[f''(x)=0\]

Answer 30

for example
\[f(x)=\sqrt[3]{x}\] has an inflection point at 0 but the derivative is not defined there

Answer 31

Not every point where the second derivative is zero will be a point of inflection, you need toi understand the overall behavior of the function in order to determine whether it is or not.

Answer 32

In this case the denominator tells you there is asymptotic behavior at 2 so you should discount a point of inflection there.

Answer 33

thank you so much for your help