Question

find the relative minimum/maximum, the point of inflection and trace the graph of the curve
f(x)=x^4-4x^3+3x^2

Answer 1

@rymdenbarn help..plz..

Answer 2

Okay so... in order to find the relative minimum/maximum you need to find the first derivative and solve for the variable. Find f'(x) and solve for x. Once you get the values for x, you need to test each interval. I'll show you what I mean:
f(x)=x^4-4x^3+3x^2
f'(x) = 4x^3 - 12x^2 +6x
0 = 4x^3 - 12x^2 +6x -- factor and solve

Answer 3

In this case, the derivative does not factor cleanly which is weird... hang on

Answer 4

use quadratic equation?

Answer 5

http://www.wolframalpha.com/input/?i=4x%5E3%2B12x%5E2-6x this is what I get

Answer 6

I suck at math so... Igs what did I do wrong again lol

Answer 7

if i remember right you take the derivative...solve for x..then plug in to the original equation.

Answer 8

the first derivative of the given is not factorable..so how to solve for x?

Answer 9

anyways for the point of inflection you have to find the second derivative, which is finding the derivative of the first derivative.
We have: f'(x)=4x^3-12x^2+6x
f''(x)=12x^2-24x+6
f''(x)=6(2x^2-4x+1)
which again does not factor cleanly
... but theoretically speaking you would find x when f''(x) = 0 as you did with the first derivative, plug in that x value into the original f(x) equation to get the y value and thus would be your point of inflection

Answer 10

I really don't know why this problem is being such a female dog