Question

On what intervals is the graph of g concave down?

Answer 1

Just a sec for more info

Answer 2

g'(x) = (-16 + x^2)/(-2 + x)

g''(x) = (x^2 - 4x - 16)/((x - 2)(x - 2))

Answer 3

x =16/(x - 4) tells the roots of g''

Answer 4

you just look for second derivative test

Answer 5

\[g'(x)=\frac{x^2-4x-16}{(x-2)(x-2)}\]
set g '(x)=0 ==> \(x^2-4x-16=0\)

Answer 6

were you given g" or you computed it
and sorry i meant to write g" instead i wrote g'

Answer 7

Apologies was away for a minute.

Answer 8

np

Answer 9

Oh! It's wrong I just realised. It's + 16, not -16. Yes, I did it myself I believe

Answer 10

so what is you function g(x) that you started with

Answer 11

Or I think?

Answer 12

I didn't the only info for g(x) I have is g(3) = 4.

Answer 13

ok i see
so you were given g'(x)=(16+x^2)/(-2+x)
and you computed g''
we don't really need g, i just asked to make sure your calculations were correct

Answer 14

I see.

Answer 15

and your g' is good
x^2-4x-16 at the top that's correct

Answer 16

now you need to solve this quadratic x^2-4x-16=0 we need sign of g''

Answer 17

The closest I could come is: -16/(x - 4) = x

Err I don't remember signs?

Answer 18

Also g'' has +16 instead of -16 my mistake

Answer 19

I don't think the 'x = ...' equation is right either.

Answer 20

no it is -16 i checked it

Answer 21

but what do you mean x=-16/(x-4)

Answer 22

That was just my attempt to find what x^2 - 4 + 16 solves to using x*y = 16, x + y = -4, which turns into 16/x = -x - 4.

Answer 23

oh no! you have quadratic equation
just use the quadratic formula

Answer 24

oops! This becomes 2 +/- 2i sqrt(3)

Answer 25

there is no i
this have real solutions only

Answer 26

there are two solution \[x=2\pm2\sqrt{5}\]

Answer 27

So from the interval is between the roots?

Answer 28

we don't really know yet
but again that's a good guess hehehe
between the root the - to +

Answer 29

I get my previous answer still from x^2 -4x + 16 (I realized it's actually + 16)

Answer 30

how did you get +?

Answer 31

I did the work as well, but when I checked it came to +

Answer 32

there won't be any concave down if it was +16
x^2-4x+16 has no real soltuon in fact x^2-4x+16>0 for any x
meaning that g is concave up everywhere but that's no the case

Answer 33

oh i thought your g' was 16+x^2/(-2+x)

Answer 34

Nah, it's -16 + x^2.

So the roots of this g'' is 2 +/- 2i sqrt(3) I think

Answer 35

I'm confused to what is what heheh.
ok so g'(x)=-16+x^2/(x-2)
then g"(x)=x^2-4x-16/(x-2)^2

Answer 36

no were only looking for real roots complex no

Answer 37

no you have g">0 for any x
so there is no concave down

Answer 38

It's all concave up then?

Answer 39

first that has critical values at x=4 and x=-4
x<-4 g' <0
-4<x<4 g'>0
x>4 g' >0
so it is an all out concave up

Answer 40

Sweet! Thank you so much. This problem in particular has taken too long so I came here. I never come disappointed!

Answer 41

no problem :)

Answer 42

Hey kid!! I don't see you use the information g(3) =4. why?

Answer 43

that just to tell you that 4 is max i guess
somehow you can see i said there two critical values

Answer 44

but that's a good question, i thought the poster would ask that too hehehe

Answer 45

or did i make a mistake ?

Answer 46

I don't know, just think of something more complicated than it is. Like this:
g(x) is undefined at x =2, hence x =2 is an retricemptote of the function.

Answer 47

g' is undefined at x=2 not necessarily g

Answer 48

if g(3) =4, we have 2 options: either g up or down like this