Question

Find all critical numbers and use the Second Derivative Test to determine all local extrema. y = x^3 - 3x^2 - 9x

Answer 1

@abb0t can u help?

Answer 2

Sure. First step is to find the derivative of the function. cAN YOU DO THAT?

Answer 3

yes

Answer 4

i got 3x^2 - 6x - 9

Answer 5

@Mertsj can u help?

Answer 6

This looks like the same question from yesterday. What happened?

Answer 7

No this is another one

Answer 8

well its the same problem but that one asked for the intervals where the graph is concave up and concave down.

Answer 9

So let's approach it this way: Look at the graph:
http://www.wolframalpha.com/input/?i=y+%3D+x^3+-+3x^2+-+9x

Answer 10

ok

Answer 11

Do you see that there is an inflection point somewhere between 0 and 2? And that to the left of that inflection point, the graph is concave downward? And to the right of that inflection point, the graph is concave upward?

Answer 12

yes the inflection point is 1, and it concaves up and to the right it concaves down

Answer 13

Do you know what concave up and concave down means?

Answer 14

yea

Answer 15

So this problem says to find the critical numbers. Do you know what they are and how to find them?

Answer 16

it shows in the second derivative whether it is positive or negative

Answer 17

Yea I know what they are , so the second derivative is 6x - 6 = 0 so the cirtical point is x = 1

Answer 18

The critical numbers are the numbers where the first derivative is 0.

Answer 19

ok

Answer 20

The reason they are called that is because at those points (where the first derivative is 0) the slope of the tangent is 0. If the slope of the tangent is 0, the tangent must be horizontal. If the tangent is horizontal, the function must have a high point or a low point there.