# Finding local minimum, local maximum or neither

• Mar 21st 2011, 01:24 AM
brumby_3
Finding local minimum, local maximum or neither
So I'm trying to find the local minimum, maximum or neither for the following function:

f(x)=(x-1)^2 * (x+3)^2

I've already found the stationary points (1,0), (-1,16), (-3,0)

So do I plug the x values into the second derivative which is 12x^2 + 24x - 4

so for example (12*1)^2 + (24*1) - 4 = 164 which means it is local minimum

(12*-1)^2 + (24*-1) - 4 = 116 which means a local minimum

and (12*-3)^2 + (24*-3) - 4 = 1,220 which is also a local minimum...

I feel this is wrong though. Can someone give me some guidance?
• Mar 21st 2011, 01:43 AM
FernandoRevilla
For $x=-3$ or $x=1$ , local minimum, for $x=-1$ , local maximum.
• Mar 21st 2011, 01:46 AM
brumby_3
Can you please show how you got those answers? Did you do the second derivative test like I did?
• Mar 21st 2011, 02:11 AM
Joanna
Quote:

Originally Posted by brumby_3

So do I plug the x values into the second derivative which is 12x^2 + 24x - 4

so for example (12*1)^2 + (24*1) - 4 = 164 which means it is local minimum

(12*-1)^2 + (24*-1) - 4 = 116 which means a local minimum

and (12*-3)^2 + (24*-3) - 4 = 1,220 which is also a local minimum...

Yes you plug those x-values into the second derivative, however you are squaring the 12 in this equation.

you are doing this: (12x)^2