Question

Indentify the open interval(s) on which the function is increasing or decreasing:
F(X)=(x^2)ln(x/2)

Answer 1

1) find the first derivative of the given function.
2) set this derivative = to 0 and solve the resulting equation for x. These are your "critical values."
3) draw a horizontal line (your x-axis" and plot the zeros ("critical values").

Example:

Answer 2

Here, with two critical values, we have 3 intervals. Choose 1 representative x-value from within each of the 3 intervals. Determine the sign of the 1st derivative at each of these representative values.

Your results? Please share a graph (use the Draw utility, below).

Answer 3

The derivative is f'(X)=2ln(X/2)X+X

Answer 4

I can't set the equation to zero?

Answer 5

0=2ln(X/2)X+X

Answer 6

Buddy: mind taking a second look at the given function and trying again to find its derivative?

\[f(x)=x^2*\ln \frac{ x }{ 2 }\]

Answer 7

Have you already recognized that this function is a product?
...or that ln (x/2) can be expanded using the rule of logs for quotients?

Answer 8

Let me try again

Answer 9

It is (log(X))/2

Answer 10

Rather than answer Yes or No:

ln (x/2) = ln x - ln 2

and if \[f(x)=x^2*\ln \frac{ x }{ 2 }\]

Answer 11

then the derivative is \[f'(x)=x^2*\frac{ d }{ dx }[\ln x-\ln 2]+[lnx-\ln2]*2x\]

Answer 12

I got the derivative is 2xlb(X)+X

Answer 13

Here I've used the product rule and the identity for the ln of a quotient. Pls go thru this and ask any questions necessary to clarify this process for yourself.

Answer 14

I mean 2xln(X)+X

Answer 15

Do I set that to zero now?

Answer 16

My result is slightly different.
We have to reconcile our two results.\[f'(x)=x+(2x)(lnx-\ln2)\]

Answer 17

I'm letting you decide whether our results are equivalent or not.
If they are, then yes, you set the first deriv. = to 0.

Answer 18

I am having trouble setting it to 0 again

Answer 19

2x(ln(X)-ln(2))+X=0

Answer 20

I have X=0 and X=1.2131

Answer 21

:(