Question

HELP!

Answer 1

Find and graph the coordinates of any local extreme points and the inflection points of the function y=x^3-12x+3

Answer 2

find the local extreme points by finding by taking the first derivative and seeing at what points it equals zero. find the points of inflection by taking the second derivative by seeing what points it equals zero. be careful though, just because the a derivative equals zero, does not always mean that there is a max or min at that point, only that the slope is 0

Answer 3

3x^2-12 would be the first derivative?

Answer 4

correct

Answer 5

if you do the first derivative correctly, you will find that there are extreme points at (-2, 19), and (2, -13).

doing the second derivative, you will find there is a point of inflection at (0,0)

Answer 6

why (0,0)?

Answer 7

once you find the first derivative y = 3x^2-12, you must take the second derivative.

which is y = 6x. to find the point of inflection, we must set the 'y' in the second derivative to 0 and find which x satisfy the equation

so if 0 = 6x, than x must == 0. now that we know that x = 0, we plug 0 in for x into the original equation.

so y = 0^3 -12(0) + 1... so y = 1... so the actual point of inflection is (0, 1)

turns out I was wrong because i did it really fast.

i'm sorry if i confused you

the point of inflection is (0,1),

not (0,0) like i said before, sorry

Answer 8

you mean (0,3)

Answer 9

oh wow..... yes, 3

i'm glad you caught that again, sorry!!

Answer 10

so that solves the whole problem?

Answer 11

that finds the local extreme points and the points of inflection, if that is the whole question