Question

the function g is given by g(x)=3x^2/e^(3x). On which of the following intervals is g increasing?
a)(-∞,0) b)(-∞,2/3) c)(0,2/3) d)(0,∞) e)(2/3,∞)

Answer 1

did you find the derivative?

Answer 2

no I'm not sure how to find the derivative of the denominator..

Answer 3

Here is the graph for reference before doing the work to find critical points...

Answer 4

You can use the product rule on this one, rewrite the e^(3x) term in the numerator first.
\[g(x) = 3x^2*e^{-3x}\]

Answer 5

The product rule gives.
\[g '(x) = \frac{ d }{ dx }g(x) = 3x^2*\frac{ d }{ dx }e^{-3x}~~+~~e^{-3x}*3\frac{ d }{ dx }x^2\]

Answer 6

Can you do those individual derivatives?

Answer 7

not to be argumentative, but if you think you will have an easier time with the quotient rule, you will have to pay for it in the end when you add up the fractions, which you must do to answer the question

Answer 8

idk wat the derivative of e^3x is

Answer 9

I changed it to use the product rule, quotient?

Answer 10

sorry i meant you need to use the quotient rule
the derivative of \(e^{3x}\) is \[3e^{3x}\] by the chain rule

Answer 11

you gotta get the numerator no matter what rule you use, so the product rule is a bad shortcut here

Answer 12

okay so g'(x)=e^3x(6x)-3x^2(3e^3x)/(e^3x)^2

Answer 13

ignore the denominator, factor the numerator

Answer 14

To each their own.

It was pretty easy this way too i think, after factoring.

\[g'(x) = 3xe^{-3x}[2-3x] =0\]

Answer 15

then drop the \(e^{-3x}\) since that is always positive, and concentrate on \[x(2-3x)\]

Answer 16

okay..then what?

Answer 17

can you figure out where \[x(2-3x)>0\) ?

Answer 18

oops
\[x(2-3x)>0\]

Answer 19

uhm -1

Answer 20

hmm your answer needs to be an interval
hint, \[y=x(2-3x)\] is parabola that opens down
it is positive between the zeros

Answer 21

uhm well wouldn't it be greater than zero at any negative number?

Answer 22

where is it zero?

Answer 23

Set the derivative = 0. This will tell you possible critical points to check for.
\[g'(x) = 3xe^{-3x}[2-3x]=0\]

This is true if:
\[3xe^{-3x}=0~~~~~or~~~~~2-3x=0\]

\[x=0~~~~~or~~~~~x=2/3\]

set up a number line, and check the intervals for inc/dec

Answer 24

Pick a number in the intervals:

(-infinity, 0) , (0, 2/3), and (2/3 , +infinity)

and test it in g '(x)
\[g '(x) >0~~~~increasing \]
\[g'(x)<0~~~~~decreasing\]

Answer 25

good?, you remember it sort of now?

Answer 26

okay so (2/3, ∞ ) it is decreasing?

Answer 27

The derivative tells you what the slope of a tangent line is at a certain point on the curve.
If the derivative is positive, the tangent has a positive slope, from the image, you can imagine the curve must be increasing.

Answer 28

& yes i can understand it better now thanks

Answer 29

i didn't calculate anything, but the graph is linked in my first post....
It is decreasing on \[(2/3, \infty)\]

but try a number in g '(x) that is less than 0, what do you get?

Answer 30

when it's less than zero, i get a positive number..

Answer 31

hmm, g '(-1) is
\[g'(-1) = 3*(-1)e^{-3*(-1)}*[2-3(-1)] = -3e^3 * (2+3)\]

Answer 32

negative times a positive ----g '(-1) is a negative number

Answer 33

oh ok i didn't plug it into the entire derivative. I disregarded a part of it. but it makes sense now. Thank you so much.

Answer 34

you're welcome... here are some things i found that may help ...

Answer 35

it is all open free stuff from the web site linked in the things..

Answer 36

okay thanks again. I appreciate it :)