- #1

navm1

- 44

- 0

## 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-2x

^{2}

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