# Thread: Help! sketching 3rd deg. equation?

1. ## Help! sketching 3rd deg. equation?

** This is seriously my 7th time creating this thread, please excuse my hastiness**

Hey there everyone, I would love for someone to help me out with this problem ASAP. Ive looked through my math book for the sugested method to do this based on example but I couldn't find it.

Sketch f(x), identify coordinates of any local min/max.

f(x) = 2-3x²+x^3

**What I did:: Well, since it's a Positive 3rd deg. there will be 2 turning points at most.. so.. in the derivative of f(x) those turning points should show up as zeros of the derivative?

f^1(x) = 1-6x+3x²

0 = [6+/- (root) 24]/6
x= 1.82 ; x= 0.2

**soo.. this is where im stuck..
do i plug the found values of x back into f(x)?
if so: f(0.2) = 2-3(0.2)^2+(0.2)^3
f(0.2) = 1.88

***Does this mean one of the local min/max coords is (0.2, 1.88)?

Thanks in advance to anyone who dares to tackle this monster

2. Hello buddy

Originally Posted by Lucidliving
** This is seriously my 7th time creating this thread, please excuse my hastiness**

Hey there everyone, I would love for someone to help me out with this problem ASAP. Ive looked through my math book for the sugested method to do this based on example but I couldn't find it.

Sketch f(x), identify coordinates of any local min/max.

f(x) = 2-3x²+x^3

**What I did:: Well, since it's a Positive 3rd deg. there will be 2 turning points at most.. so.. in the derivative of f(x) those turning points should show up as zeros of the derivative?
Sounds good.

Originally Posted by Lucidliving
f^1(x) = 1-6x+3x²
Where does the '1' come from?

$f'(x) = 3x^2-6x$

because the derivative of 2 is 0;
Originally Posted by Lucidliving
0 = [6+/- (root) 24]/6
x= 1.82 ; x= 0.2
That's wrong anyway

Consider
$f'(x) = 3x^2-6x = x(3x-6) = 0$

$\Rightarrow$ x = 0 and x = 2

Originally Posted by Lucidliving
**soo.. this is where im stuck..
do i plug the found values of x back into f(x)?
if so: f(0.2) = 2-3(0.2)^2+(0.2)^3
f(0.2) = 1.88
Yes, you do. Because you want to find the x (you already know them) and y-coordinates.
Originally Posted by Lucidliving
***Does this mean one of the local min/max coords is (0.2, 1.88)?

Thanks in advance to anyone who dares to tackle this monster
You have to find it out by considering the second derivative at x = 0 and x=2

Is
f''(0) > 0 => min

f'(0) < 0 => max

f''(2) > 0 => min

f''(2) < 0 => max

Yours
Rapha

3. Thank you so much! !!!!
!!!
!!!!!

Im so glad there are people like you willing to help for absolutely no payment of any kind. Except perhaps a social incentive.

I wish I could give back.. but thats why Im here, I cant.. =/
Unless someone wants to know about Physics or Economics!