Question

Consider the function f(x) = x^{2}e^{3 x}.
For this function there are three important intervals: (-\infty, A], [A,B], and [B,\infty) where A and B are the critical numbers.
Find A
and B
For each of the following intervals, tell whether f(x) is increasing (type in INC) or decreasing (type in DEC).
(-\infty, A]:
[A,B]:
[B,\infty):

Answer 1

A function is Increasing when the slope is positive or f'(x) >0
A function is Decreasing when the slope is negative or f'(x) <0

The critical points occur when slope is zero ... f'(x) = 0

so step 1 : take derivative of function

Answer 2

use product rule:
\[(fg)' = f'g + fg'\]
where f = x^2 and g = e^3x

Answer 3

uh huh

Answer 4

So what did you get for the derivative?

Answer 5

f'(x)=2xe^3x+x^2e^3x(3)

Answer 6

Yes... now factor out the greatest common factor in each term and set the whole thing equal to zero.

Answer 7

e^3x(2x+3x^2)=0

Answer 8

x comes out too

Answer 9

\[e^{3x}x(2+3x)=0\]

Answer 10

e^3x will never equal zero... but that x right next to it will...
so xe^(3x)=0 and 2+x=0 solve for x in each and you will have A and B

Answer 11

I dont know what I am doing

Answer 12

You seem to know a bit... work from there. correction above... you are right... the term on the right it (2+3x), not 2+x. See... you know better than me! :)

Answer 13

haha I know how to do the derivative thats about it

Answer 14

and factor apparently :)

Answer 15

Ohh i find the critical numbers

Answer 16

OKay let me see.................

Answer 17

so when two things are multiplied and equal to zero, either factor can make the whole thing zero.... yes this is how you find the crits

Answer 18

Soi s one number 0

Answer 19

is*

Answer 20

yes

Answer 21

The other comes from 2+3x=0, because e^(3x) can never be zero.

Answer 22

-2/3

Answer 23

?

Answer 24

yes

Answer 25

woo hoo

Answer 26

woof!

Answer 27

Ha okay I dont know what to do next

Answer 28

What is A and B

Answer 29

Cause I put in 0 and -2/3 and it said it was wrong

Answer 30

Now I draw a number line with -2/3 and 0 on it.
Find a test point in each interval that they specify and see if the derivative is positive or negative in those intervals... I used -1, -1/3, and 1.

It is -2/3 first... then 0 for your program.

Answer 31

ooh heh

Answer 32

\[(-\infty, - \frac{2}{3}) U (- \frac{2}{3}, 0) U ( 0, \infty)\]

Answer 33

Oh yes I just test those points to see right?

Answer 34

I get it!!

Answer 35

Thank you!

Answer 36

Don't test the crits... you'll get zero... pick something to the sides of them that are easy to calculate.

Answer 37

Yes, I knw wher to test

Answer 38

Cool!

Answer 39

Its INC DEC INC

Answer 40

That's what I got! :)

Answer 41

woof! :)

Answer 42

Can you help me with something else?

Answer 43

Post a new question... then I'll look at it.

Answer 44

okay