Question

PLEASEEE HELPP ME! f(x) = 9x4 + 5x3 - 4x2 - 7x - 1 what are the exact two inflection points? Kay thanks (:

Answer 1

derive it twice and solve for 0; that will give you points to test

Answer 2

also where it is undefined

Answer 3

36x^3+15x^2-8x-7

Answer 4

good; thats one derivative :)

Answer 5

108x^2+30x-8

Answer 6

nice work; now we either factor it or use the quadratic formula to determine the "zeros"

Answer 7

This is the part I can never understand at all! The solve for 0 part

Answer 8

by solving for "0" we are determiningg the places where the slope is a horizontal line

Answer 9

at those places we tend to have critical values

Answer 10

http://www.wolframalpha.com/input/?i=108x^2%2B30x-8
to check the roots, wolfram gives us these options

Answer 11

this version might help in "seeing" what the derivatives look like

Answer 12

http://www.wolframalpha.com/input/?i=+y%3D9x4+%2B+5x3+-+4x2+-+7x+-+1+and+y%3D36x^3%2B15x^2-8x-7+and+y%3D108x^2%2B30x-8

Answer 13

ohh so. 108x^2+30x-8=0 so two values that added together are -8 and multiplied are 30?

Answer 14

how do I get the answer from that graph :/

Answer 15

??????

Answer 16

the zeroes for this might be better found with the quadratic formula such that:

-30 +- sqrt(30^2 -4(108)(-8))
x = ---------------------------
2(108)



-30 +- sqrt(900 +108(32))
x = ---------------------------
2(108)

im just gonna google calculate it ....

(-30 + sqrt(900 +108(32)))/(2*108) = 1/6
(-30 - sqrt(900 +108(32)))/(2*108) = -4/9

Answer 17

so when x=1/6 ; or x=-4/9, we have a critical value to test

Answer 18

becasue of the nature of the original equation; its a 4th degree, I already know these are the inflection points :)

Answer 19

Ohhh emmm geee! Thanksssss! You're the best!

Answer 20

youre welcome