# Help with finding stationary point (work needs checking)

• Aug 10th 2010, 10:38 PM
99.95
Help with finding stationary point (work needs checking)
Let f( x ) = 2x^3 - 3x^2 - 12x + 6, x E R (E= element of). Then (give the exact coordinates and the type of each stationary point such as local maximum, local minimum, and stationary point of inflexion)

Here is my working, please notify me where i've made errors and how to fix them with correct procedure/calculation.
f( x ) = 2x^3 - 3x^2 - 12x + 6, x E R
f'(x) = 6x^2 - 2x - 12
Let f'(x)= 0
Therefore 6x^2-2x-12=0
a = 6 b = -2 c= -12
Results:
[see attached image]

f( result 1 ) =
f( result 2 ) =
(inability to use latex, i shall upload my answers shortly, also i'm having difficult onwards after factorise 6x^2-2x-12=0)
Thanks.
• Aug 10th 2010, 10:51 PM
yeKciM
Quote:

Originally Posted by 99.95
Let f( x ) = 2x^3 - 3x^2 - 12x + 6, x E R (E= element of). Then (give the exact coordinates and the type of each stationary point such as local maximum, local minimum, and stationary point of inflexion)

Here is my working, please notify me where i've made errors and how to fix them with correct procedure/calculation.
f( x ) = 2x^3 - 3x^2 - 12x + 6, x E R
f'(x) = 6x^2 - 2x - 12
Let f'(x)= 0
Therefore 6x^2-2x-12=0
a = 6 b = -2 c= -12
Results:
[see attached image]

f( result 1 ) =
f( result 2 ) =
(inability to use latex, i shall upload my answers shortly, also i'm having difficult onwards after factorise 6x^2-2x-12=0)
Thanks.

$\displaystyle 6x^2-6x-12=0$

u can't do this :

$\displaystyle \displaystyle \frac {d}{dx} 3x^2 = 2x$ ?????

result will be :
$\displaystyle x_1=-1 \; \; ; x_2=2$
• Aug 10th 2010, 10:59 PM
99.95
silly mistakes will always be my downfall and thank you very much