Question

Please help with this calculus problem!



f(x) = x / (x^2 - 9) I need to sketch this function without using technology and I need to find:
i) Extremas
ii) Points of inflection
iii) Concavities

Answer 1

Hi Turing :)

Answer 2

1)
extrema occur when f'(x)=0
as we discussed before

Answer 3

Hi :)

Answer 4

Ok, for the derivative, I got:
\[-x ^{2}+9/(x ^{2}-9)^{2}\]

Answer 5

I have no idea on how to solve for x in this one

Answer 6

I don't think that's right
I'm sorry I have to go....
work on that derivative: use the product rule and chain rule

Answer 7

Ok, bye

Answer 8

AccessDenied can you help?

Answer 9

yeah, im just going over the problem myself to make sure i have it straight

Answer 10

Ok

Answer 11

rewrite it as\[x(x^2-9)^{-1}\]and use the product rule and chain rule
(quotient rule is for suckers!)

Answer 12

LOL thats how I did it (quotient rule)

Answer 13

product rule is easier in many cases, as you will see

Answer 14

i used quotient rule too, even though i normally do product rule with negative exp
:(

Answer 15

you can derive the quotient rule from the product rule in about 2 seconds

Answer 16

\[(uv^{-1})'=u'v^{-1}-uv^{-2}v'=v^{-2}(u'v-v'u)=\frac{u'v-v'u}{v^2}\]therefor the Q rule is a waste of time in my opinion, but to each their own :)

Answer 17

:'(

Answer 18

I should write\[(\frac uv)'=(uv^{-1})'=u'v^{-1}-uv^{-2}v'=v^{-2}(u'v-v'u)=\frac{u'v-v'u}{v^2}\]

Answer 19

Are you sure that my derivative is wrong?

Answer 20

no I didn't check :P
just looked wrong
like I said gotta go!

Answer 21

OK, Bye

Answer 22

its pretty close
it might just be what you're writing to show it

Answer 23

like, do you mean for it to be this?

\[ -\frac{x^2 + 9}{(x^2 - 9)^2} \]

Answer 24

Thats exactly what I got

Answer 25

does it make a difference if I apply the - sign to the x?

Answer 26

yeah, then its correct... you just had it written in a not-so-obvious way. :P

Answer 27

so its -x^2?

Answer 28

the "-" at the outside would go through to both the x^2 and 9

\[ \frac{-x^2 - 9}{(x^2 - 9)^2} \]

Answer 29

Oooohh

Answer 30

Ok, now that we have the derivative, how do we solve for x?

Answer 31

Can you also include steps :P

Answer 32

we have to set both the numerator and the denominator = 0 and try to get the x-value
(if the numerator = 0, then the expression is 0; and if the denominator is 0, then the expression is undefined)

-x^2 - 9 = 0

(x^2 - 9)^2 = 0

Answer 33

I got 3 for the first one, I dont know how to do the second one

Answer 34

the first one, you wouldn't have any real solutions.
-x^2 - 9 = 0
-x^2 = 9
x^2 = -9 <-- no real solution to that!

Answer 35

Oh lol

Answer 36

the second one actually does have solutions.
(x^2 - 9)^2 = 0
sqrt( (x^2 - 9)^2 ) = sqrt( 0 )
|x^2 - 9| = 0 absolute values not needed because |a| = 0 <=> a=0
x^2 - 9 = 0
x^2 = 9
sqrt( x^2 ) = sqrt( 9 )
|x| = 3
x = +- 3

Answer 37

Ok I get that

Answer 38

except, to be extrema, the point has to be on the graph of the original function.
x=3 and x=-3 are both undefined for the original function as well. So, we'll have no extrema for the graph.

Answer 39

Ok

Answer 40

So, do you remember how to find the points of inflection?

Answer 41

http://ocw.mit.edu/courses/mathematics/18-01-single-variable-calculus-fall-2006/video-lectures/lecture-10-curve-sketching/

Answer 42

You need to find the concavities

Answer 43

You find the second derivative and the inflection is at that point., right?

Answer 44

the points where the second derivative is 0 or undefined, and exists for the first derivative as well.

Answer 45

So, there is no point of inflection?

Answer 46

have you tested the second derivative?

Answer 47

No

Answer 48

itd be good to check the second derivative first before you assume it doesnt! :)

Answer 49

I dont know how to take the derivative of this function

Answer 50

okay, i gotta go through it a second time myself. its not easy to do, but we can use the same technique as the first derivative. There's just more going on this time. :P

Answer 51

Yeah, I already made so many mistakes :/

Answer 52

heres what wolfram says: http://www.wolframalpha.com/input/?i=derivative+of+-%28x%5E2+%2B+9%29%2F%28x%5E2+-+9%29%5E2

yeah, i keep making mistakes too. I need to try one more time!

Answer 53

This is very confusing

Answer 54

I cant get this second derivative...

Answer 55

okay, i finally got it after realizing that i could multiply things together for magic

i'll link my work, if it possibly helps

Answer 56

That is very difficult to read, but atleast you got the right answer :P

Answer 57

i used the product rule with negative exponents

i changed it to exponent form on the first equals-sign
second, i used product rule
then, i brought the negative exponents back into standard
i got common denominators on the first term with the second by multiplying both top and bottom by (x^2 - 9). I also multiplied the -4x through
i then multiplied the 2x through so that i could brng the fractions together and add like-terms
once i had it down in that form, i just factored out a 2x and removed the negative.

Answer 58

OK=k

Answer 59

Hold on, im still trying to read your work :P

Answer 60

i could probably rewrite it in latex, it'd just take another minute. lol

Answer 61

On the fourth = sign, that line, whats the denominator for the first fraction?

Answer 62

the exponent on that term

Answer 63

should be "(x^2 - 9)^3"

Answer 64

ok

Answer 65

on the last line its -2x^3-5 and then what

Answer 66

-2x^3 - 54x

Answer 67

ok I got it now

Answer 68

So what do we do with this second derivative

Answer 69

we want to find the points where it is 0 or undefined, where the first derivative does exist.

It has the same points where it is undefined as the first derivative, but we have a new 0. The numerator has "2x" as one term, which when x=0, makes the numerator 0.

Answer 70

So this is undefined aswell right?

Answer 71

the second derivative is undefined for x=3 and x=-3, but since they don't exist on our first derivative either, its not a inflection point.

Answer 72

So what should I write for the inflection point?

Answer 73

the inflection point is at x=0 since that is where the second derivative is 0.

Answer 74

Ok

Answer 75

So we have to equate the second derivative to 0 but when we do that we get the numerator at 0 right?

Answer 76

Am I right?

Answer 77

AccessDenied, are you there?

Answer 78

yes, set the numerator equal to zero

Answer 79

Alright now that I know that the inflection point is at zero, and I think we've determined that there are no concavities we just need to find the extremas

Answer 80

concavity is from testing the intervals around the inflection points.

sorry, i had to reply to somebody else about stuff. D:

Answer 81

So what did we determine from taking the first derivative?

Answer 82

the first derivative was to find if there were any extrema
we check if the x-values that make the second derivative 0 or undefined in the first derivative to make sure that they are on the graph since otherwise, it wouldn't be an inflection point.

Answer 83

And we determined that there were no concavities right?

Answer 84

hmm.. not sure how we should test for this one. seems like the concavity changes between the x=3 and x=-3 as well...

Answer 85

maybe TuringTest is more familiar with it. D:

Answer 86

You said that +-3 was undefined for the original function

Answer 87

@TuringTest could you explain this? I'm definitely not remembering how this would be done. >.<

Answer 88

First derivative set equal to zero = extreamas.
Second derivative set equal to zero = points of inflection

Answer 89

you just pick a point in the interval
the concavity changes at \(x=\pm3\) so we get the intervals\[(-\infty,-3)(-3,3)(3,\infty)\]so check a point in each interval, and plug that point into \(f''(x)\)
if \(f''<0\) it is concave down
if \(f''>0\) it is concave up

Answer 90

Isnt +-3 undefined for the original function though?

Answer 91

so?
that can still be a point where the concavity changes

Answer 92

Oh

Answer 93

OK

Answer 94

they are asymptotes
that often changes concavity from one side to the other

Answer 95

Ok, so how should I write the answer for the concavities?

Answer 96

so, are they also considered "inflection points"?

Answer 97

http://www.wolframalpha.com/input/?i=plot%20f(x)%20%3D%20x%20%2F%20(x%5E2%20-%209)&t=crmtb01
notice how the concavity changes between x=-3 and x=3

the inflection points...