Question

Find the points of inflection for the y=x^5-15x^3

Answer 1

take the derivative twice. set the result = 0 and solve

Answer 2

y'=5x^4-45x^2

y''=20x^3-90x
0 =10x(2x^2-9)
\[x=0, x=\pm \sqrt{\frac{9}{2}}\]
so are my intervals going to be

Answer 3

\[y=x^5-15x^3\]
\[y'=5x^4-45x^2\]
\[y''=20x^3-90x\]

Answer 4

oh looks like you got it yes?

Answer 5

\[10x(2x^2-9)=0\]
\[x=0\]
\[x=\pm\sqrt{\frac{9}{2}}\]

Answer 6

forget sign chart

Answer 7

i cant i have to do it that way...lol

Answer 8

you have a third degree polynomial with positive leading coefficient yes?

Answer 9

hell no. think of what the picture looks like. you have 3 zeros
\[a,b,c\] and a third degree polynomial. it will be negative on
\[(\infty,a)\]

Answer 10

positive on
\[(a,b)\] negative of
\[(b,c)\] and positive on
\[(c,\infty)\]

Answer 11

Do i use all those intervals in my sign chart

\[(-\infty,-\sqrt{\frac{9}{2}}), (-\sqrt{\frac{9}{2}},0), (0,\sqrt{\frac{9}{2}}), (\sqrt{\frac{9}{2}},\infty)\]

Answer 12

yes and it will be
negative
positive
negative
positive
respectively

Answer 13

i have to use all of 'em?

Answer 14

making your original 5th degree poly
concave down
concave up
concave down
concave up

Answer 15

yes you have 4 intervals

Answer 16

lol okay thanks satellite!!!

Answer 17

yw. btw throw out that "sign chart" it is a time waster. just remember what a third degree poly looks like and use common sense

Answer 18

for the 2nd interval i got a negative

Answer 19

my lecturer says we have to use it lol...

Answer 20

oh nvm no i didnt

Answer 21

tell him/her not to waste your time. why do they bother to teach what a third degree poly looks like if you cannot use it later?

Answer 22

lol idk

Answer 23

what a nice picture
http://www2.wolframalpha.com/input/?i=y%3D20x^3-90x

Answer 24

as you can see....
negative
positive
negative
positive
sign chart be damned

Answer 25

Refer to the attachment

Answer 26

okay thanks satellite

Answer 27

oh wow thanks robtobey!

Answer 28

Your welcome.