# Homework Help: Finding stationary points. no x in second derivative

1. Mar 27, 2014

### navm1

1. The problem statement, all variables and given/known data
Just started getting introduced to calculus and a couple applications. After i've found the stationary point i understand that i can put the x value into the second derivative to find if its a maximum or minimum point. i.e

12x-2x2
ƒ'(x)= 12-4x
12-4x=0
x=3
so 12(3)-2(3)2 = 18
So if i've worked it out correct then the stationary point is 3,18

3. The attempt at a solution
So if i take the second derivative i have

ƒ''(x)= -4

if theres no x in the second derivative then how do I find out whether it's a maximum or minimum value?

Thanks

2. Mar 27, 2014

### SteamKing

Staff Emeritus
You can always plot the original function.

3. Mar 27, 2014

### Staff: Mentor

You can use the 2nd derivative to determine if a critical point is a local maximum or minimum. Your textbook should have this test and some examples. Also, y = 12x - 2x2 is a very simple function. A quick sketch of its graph, as SteamKing suggests, will show immediately what's going on.

4. Mar 27, 2014

### LCKurtz

You ask yourself whether $-4$ is positive or negative for concavity.