Finding stationary points. no x in second derivative

[navm1]

Homework Statement

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

The Attempt at a Solution

So if i take the second derivative i have

ƒ''(x)= -4

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

Thanks

 

[SteamKing]

You can always plot the original function.

 

[Mark44]

Homework Statement

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

The Attempt at a Solution

So if i take the second derivative i have

ƒ''(x)= -4

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

Thanks

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.

 

[LCKurtz]

So if i take the second derivative i have

ƒ''(x)= -4

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

Thanks

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