Question

Question about function analysis

Answer 1

\[f(x)=\frac{ 1 }{ 5 }x^5-\frac{ 8 }{ 3 }x^3+16x\]

Answer 2

ok so I know this is all real numbers

Answer 3

the y intercept is 0

Answer 4

somewhat unsure why the x intercept is 0

Answer 5

I've analyzed the function, and it is indeed a function of x.

Answer 6

i know its an odd fuction and symetrical about the origin

Answer 7

no asymptotes

Answer 8

when i do the first derivative test, i get

Answer 9

\[f'x=5x^4-8x^2+16\]

Answer 10

the book says there are no critical points, if that because i cannot factor the first derivative?

Answer 11

in short, anytime you have a polynomial function, domain is all reals, no asymptotes.
as far a critical points, max/mins you'll have to do the calculus on the function.

Answer 12

d/dx (x^5/5) = 5/5 x^4 = x^4

Answer 13

!

Answer 14

to get x-intercept, put f(x) = 0

Answer 15

how embarrassing

Answer 16

>:0

Answer 17

x^4-8x^2-16 =0 -->x=+2 or -2

Answer 18

so the interval for inc/dec is

Answer 19

increase\[(-\infty,\infty)\]

Answer 20

because -2, and 2 will yield increase even between them will be increase :D

Answer 21

THANKS @hartnn !! :D

Answer 22

there isnt a local min or max because the +- 2 arent critical points?

Answer 23

u got x=+/-2 when u equate f'(x)=0
so aren't those critical points ?

Answer 24

and graph does change its slope at x=+/-2

Answer 25

http://www.wolframalpha.com/input/?i=+f%28x%29+%3D+x%5E5%2F5+-%288%2F3%29x%5E3%2B16x

Answer 26

ahh, so i do need those, they change curvature at +-2 , not inc/dec

Answer 27

the interval for increasing was correct (-infinity , infinity)

Answer 28

yep ;D

Answer 29

to find inflection points, i just plug in the critical points from the 2nd derivative to the origional function?

Answer 30

IP=\[-2,\frac{ -256 }{ 15 }\]

Answer 31

\[2,\frac{ 256 }{ 15 }\]

Answer 32

\[0,0\]

Answer 33

all IPS :D

Answer 34

thanks for making me revise what are inflection points :D

Answer 35

YAY!!! i helped @hartnn kinda!

Answer 36

yes you did!

Answer 37

and those IP's are correct :)

Answer 38

Thanks again!!! hart i may have some more qs, later :)

Answer 39

okay , if i'll be online, i'll be there.