1. ## Finding relative extremas

Find all relative extremas. Use the Second Derivative Test when applicable. f(x)=(x^3)-(3x^2)+3... What I did:f'(x)=(3x^2)-6x 0=(3x^2)-6x -> 6x=3x^2 -> 6x=x(3x)6=3x -> x=2, 0, f''(x)=6x-60=6x-6 -> 6=6x -> x=1 After getting the critical numbers, how do I get the relative extremas again? Through the sign chart I think, but I'm a bit confused how to set that sign chart up. Answers: Relative minimum: (0, 3)Relative maximum: (2, -1)

2. ## Re: Finding relative extremas

At points where f'(x) = 0, these points are called the critical points, meaning that the function at those points are either relative minimums, relative maximums, or saddle points. So You got x = 2, 0 as the critical points. Now to determine if these critical points are local minimna or local max, you can apply the second derivative test.

if f''(2) < 0 then at the point x = 2, f is a relative maxima,
if f''(2) > 0 then at the point x = 2, f is a relative minima.
if f''(2) = 0, the test won't work and u need some other method of determining if relative max or min

Same for the point x = 0

3. ## Re: Finding relative extremas

First a note on wording: Extrema is the plural of extremum, so you would say you are finding all relative extrema, which include all minima and maxima. At least that's what I was taught many years ago.

We are given:

$f(x)=x^3-3x^2+3$

You have correctly differentiated to find:

$f'(x)=3x^2-6x$

Now, we want to equate this to zero and find all roots:

$f'(x)=3x(x-2)=0$

We see our critical numbers are:

$x=0,\,2$

I see you found these, but I was unable to really follow your working.

Now, you want to find the second derivative (which I think you did):

$f''(x)=6x-6=6(x-1)$

Evaluate the second derivative at your 2 critical numbers, and recall if:

$f''(c)<0$ then the critical number is at a local maximum.

$f''(c)>0$ then the critical number is at a local minimum.

4. ## Re: Finding relative extremas

Oh!!!
I see now!
So you plug in the critical numbers for 'c' in the second derivative right?
And if one of them follows those rules, then it's the maxima or minima right?

Ok thanks!
I was a bit confused on the second derivative test at first cause of 'c' xD
So 'c' is the critical numbers you found right?

5. ## Re: Finding relative extremas

Yes c represents the critical number(s), which you test separately. Also, I should add that if you find the second derivative is zero then there is no relative extremum at that point.

6. ## Re: Finding relative extremas

Originally Posted by MarkFL2
Yes c represents the critical number(s), which you test separately. Also, I should add that if you find the second derivative is zero then there is no relative extremum at that point.
Hmm, I'm a bit confused
So is this what you do for the second derivative test.
After you get your critical numbers, which is 0, and 2
So the second derivative is f''(x)=6x-6 = 6(x-1)
So if I plug in f''(0)=6(0)-6 = -6
f''(2)=6(2)-6=12

OR

f''(0) = 6(0-1) = -6
f''(2) = 6(2-1) = 6

After seeing they're both -6 and 6, what does that mean to find the relative minimum and relative maximum o.o

The answers are: Relative minimum: (0, 3), Relative maximum: (2, -1)

7. ## Re: Finding relative extremas

We have the two roots of the first derivative as critical numbers $x=0,\,2$.

So, using the second derivative $f''(x)=6(x-1)$ we evaluate this second derivative at the critical numbers:

i) $f''(0)=6(0-1)=-6$

Now, since this is less than zero, we know we have a relative maximum at:

$(0,f(0))=(0,3)$

ii) $f''(2)=6(2-1)=6$

Now, since this is greater than zero, we know we have a relative minimum at:

$(2,f(2))=(2,-1)$

8. ## Re: Finding relative extremas

Originally Posted by MarkFL2
We have the two roots of the first derivative as critical numbers $x=0,\,2$.

So, using the second derivative $f''(x)=6(x-1)$ we evaluate this second derivative at the critical numbers:

i) $f''(0)=6(0-1)=-6$

Now, since this is less than zero, we know we have a relative maximum at:

$(0,f(0))=(0,3)$

ii) $f''(2)=6(2-1)=6$

Now, since this is greater than zero, we know we have a relative minimum at:

$(2,f(2))=(2,-1)$
Oh!!!
Thank you, now I get it!
So after you see what fits into the second derivative test, you plug the critical points back into the original equation!

Now that makes more sense!