Question

Find the relative extrema for f(x); be sure to label each as a maximum or minimum. You do not need to find function values; just find the x-values.

Answer 1

So do I just look for the x values?

Answer 2

So we need to find x values that we can classify as critical points or extrema.
we'll need to take the derivative of our function f.

Do you understand how to do that?

Answer 3

yah let me show u what i think.....

Answer 4

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

Answer 5

f"(x)=9x^2-6x.....

Answer 6

9x^2-6x=0

Answer 7

Hmm looks like you're missing a term in f'(x)... what happened to the 6x when you took it's derivative?

Answer 8

f"(x)=9x^2-6x-0?????

Answer 9

x=4/3,0?

Answer 10

nope

Answer 11

x=2/3 and 0

Answer 12

\[\Large\bf f(x)\quad=\quad \frac{3}{4}x^4-x^3-3x^2+6x\]
\[\Large\bf f'(x)\quad=\quad 3x^3-3x^2-6x\color{red}{+6}\]You're missing a term in your first derivative. See it in red here?

Answer 13

ok

Answer 14

f"(x)=9x^2-6x-6?????

Answer 15

We don't need the second derivative.
The question is specifically asking for max/min points.
So we only need to deal with the first derivative.

Our next step is to set the derivative function equal to zero and solve for x.\[\Large\bf 0\quad=\quad 3x^3-3x^2-6x+6\]

Answer 16

ok one sec

Answer 17

x=sqrt2,-sqrt2, and 1.

Answer 18

Mmmm ya that sounds right!

Answer 19

now what?

Answer 20

Err ya I guess that might make more sense :P
Just plug the x values back into the original function, the one that gives you the largest output in your max.

And the smallest output your min.

Answer 21

\[\Large\bf f(-\sqrt2)=?\]\[\Large\bf f(1)=?\]\[\Large\bf f(\sqrt2)=?\]

Answer 22

ok thnxs could u help me with another one?

Answer 23

This is for the same function?
Oh darn, I shouldn't had erased the picture I was drawing lol.

Answer 24

yah same function

Answer 25

Let's see what's happening on the left side of our point x=-sqrt2

Answer 26

\[\Large\bf f'(-2)=?\]We only care about the `sign` of the result, is it negative or positive?

Answer 27

-?

Answer 28

Mmm ya that sounds right!So we're decreasing in this interval, from negative infinity to -sqrt2.