Question

I'm studying for a final in calculus that is cumulative. On one of my old exams, I had the following question...
Find all relative extrema of the function
f(x)=x^3-6x^2+15
My answer written out was
f(x)=x^3-6x^2+15
f'(x)=3x^2-12x
3x^2-12x=0
3x(x-4)=0
x=0,x=4
f(0)=15 f(4)=-17
Final Answer rel max=15
rel min=-17
The teacher wrote "Why? and took 4 points off. what was I missing?

Answer 1

i guess you are supposedto explain that is is a relative max because...

a) the derivative changes sign from positive to negative or
b) the second derivative is negative at that point and so it is concave down, so you have a max.

Answer 2

Ah I always had problems with concavity and was never sure where I needed to take the 2nd derivative

Answer 3

or course the real explanation is that you have a cubic with positive leading coefficient and htree critical points, and so since you are not an idiot you know it look like this

Answer 4

Yeah

Answer 5

It's a good habit to think of maximums as points where the derivative turns from positive to negative. It ties into the definition of derivative as rate of change, and much easier to remember than concavity IMO.

Answer 6

Thanks!

Answer 7

so for that would I still need the 2nd derivative?

Answer 8

Depends on what the derivative looks like. For a quadratic you can see it by just looking at it.

Answer 9

Thanks, but I think satellite deserved the medal more ;)

Answer 10

I'm brand spankin' new at this forum. Both were very helpful and I always like to come at problems from a different angle.

Answer 11

Cool. I'm not complaining either.

Answer 12

:-)