Question

f(x) = x + (128) / (x^2+16). Identify all relative extrema.

Can anyone help me?

Answer 1

r u in calculus?

Answer 2

yes?

Answer 3

okay. sorry
find f'(x)
then set f'(x) = 0

Answer 4

Alright.

Answer 5

Ok So it would be ((254x) / (x^2 + 16)^2) = 0

Answer 6

This is the point where I originally got, but what do I do once it equals zero?

Answer 7

isn't f'(x) = 1-128/(x^2+16)^2 ?

Answer 8

I am not sure.. I got ((1 - 254x) / (x^2 + 16)^2) = 0

Answer 9

how?
can u show ur work?

Answer 10

Could you tell me first what I would do once I make it equal to 0?

Answer 11

wait. ur f'(x) is wrong.

Answer 12

oh.

Answer 13

Try it this way by rewriting, then finding derivative
\[x+128(x ^{2}+16)^{-1}\]

Answer 14

Alright well 1-128/(x^2+16)^2 .. What do I do next?

Answer 15

Let\[x ^{2}+16\]be u, find derivative of u

Answer 16

Alright guys, I know how to derive it, the problem is actually finding the relative extrema. So I let the derivative equal zero..then what?

Answer 17

1-128/(x^2+16)^2 = 0
u'll have to do some algebra here (common denominator...)

Answer 18

The problem is, we don't agree that you have correct derivative of the thingie.

Answer 19

well if I knew what I do when it equals zero I can figure out if it's really right or wrong.

Answer 20

You have to solve for x.

Answer 21

oh.

Answer 22

I redid it and I got 1 - 128(2x) / (x^2 + 16)^2

Answer 23

In the answer choices it says that it is either 1.182, and 4 or -1.182, and -4

Answer 24

first u need to find common denominator before solving for
1 - 128(2x) / (x^2 + 16)^2 = 0

Answer 25

not really sure about this one, would it be (x^2 + 16)^2?

Answer 26

yeah. but the 2x shouldn't be there.

Answer 27

oh I see, so when everything equals to 0, what exactly am I supposed to be doing? This part always confuses me.

Answer 28

it should just be 1 - 128/ (x^2 + 16)^2 = 0

Answer 29

It's been a while since I did derivatives so I punched it in Wolfram
\[1-\frac{ 256x }{ (x ^{2}+16)^{2} }\]

Answer 30

That's what I was trying to say the derivative was^

Answer 31

Alright So what would I do when it equals 0.. Can you give me the results/process of how to do it.

Answer 32

(x^2+16)^2-128 = 0
we don't really care about the denominator because if the numerator equals 0 then the equation equals 0

Answer 33

Shouldn't 128 by 256x?

Answer 34

Shouldn't it 'be'

Answer 35

yea I was saying that earlier.. because of when you use the quotient rule.

Answer 36

oh wait sorry yeah it should. i was thinking something else.
okay so f'(x) = 1-256x/(x^2+16)^2
so then set it = 0
(x^2+16)^2-256x = 0
and solve.

Answer 37

Hmm..How would I go about solving this? Since I am solving for x do I have to expand (x^2 + 16)^2?

Answer 38

Multiply each term by (x^2 + 16)^2 to remove fraction

Answer 39

yeah you have to expand (x^2+16)^2

Answer 40

Alright so I have x^4 + 16x^4 + 256 -256x = 0 at the moment

Answer 41

any idea what I should focus on next?

Answer 42

chag?

Answer 43

My algebra might be wrong, I got x=0, -32, -256. If it is, or if it were right you would do something like this