Question

Find the local maximum and minimum values of f using both the First and Second Derivative Tests.

f(x)=x/(x^2+16)

Answer 1

take first derivative, set it equal to 0, solve for x. also find where the first derivative is udnefined..

Answer 2

find all of those x values and then take 2nd derivatives..

Answer 3

plug ur values into the 2nd derivative and whenever u get negative then u have max, and positive = min

Answer 4

+4 and -4

Answer 5

No the derivative, not the actual thing

Answer 6

\[\frac{d}{dx}\frac{x}{x^2+16}=\frac{16-x^2}{(x+16)^2}\]
zeros are at 4 and-4

Answer 7

oh okay so she was right.. and als x = -16 too

Answer 8

??

Answer 9

oh no that is not right

Answer 10

-4 is a min, 4 is a max sorry

Answer 11

nothing like revisionist history

Answer 12

do you really need second derivative test here? it is clear as day that
\[(x+16)^2>0\] for all x, so only need sign of
\[16-x^2\] which is obviously negative on
\[(-\infty,-4)\cup(4,\infty)\] and positive on
\[(-4,4)\] making -4 a local min and 4 a local max

Answer 13

Satellite you missed x = -16 as a local extremum!

Answer 14

but you can use it if you want. the second derivative is
\[\frac{2x(x^2-48)}{(x+16)^2}\]

Answer 15

Satellite -16 is another extremum u should test lol sorry if im annoying u!

Answer 16

hold the phone. why is -16 an extremum? i mean i believe you i just don't see it?

Answer 17

In the denominator:
(x+16)^2 there are extrema when 1st derivative is 0 or when it doesn't exist

Answer 18

At x = -16 denominator is 0, there's a cusp

Answer 19

derivative is
\[\frac{16-x^2}{(x^2+16)^2}\] yes?

Answer 20

the denominator is always positive

Answer 21

in fact it is always greater than 16

Answer 22

ohh wait is it? eariler u said it was just x not x^2

Answer 23

let me look at the problem again

Answer 24

I didn't take the derivative, sorry, no i meant according to ur 2nd or something post it would've been that..

Answer 25

oh damn. i made a typo twice sorry. it is
\[\frac{d}{dx}\frac{x}{x^2+16}=\frac{16-x^2}{(x^2+16)^2}\]

Answer 26

my mistake

Answer 27

i forgot the square on the x

Answer 28

Oh okay, then you are good..

Answer 29

and second derivative is
\[\frac{2x(x^2-48)}{(x^2+16)^2}\]

Answer 30

sorry that was my fault

Answer 31

extrema are correct though

Answer 32

yup

Answer 33

so is the answer -1/8 and 1/8

Answer 34

are you there?