1. ## function g(x)=(1-2x)^2(x-3)

g(x)=(1-2x)^2(x-3)
I need to state whether the function is odd or even and i said odd since there is a degree of 3, and the x and y intercepts i solved for and ended up with y=-2 and x=3,1/2. now i need to find the first and second derivative, and i was wondering if i should use the chain rule and product rule ?

2. I tried solving for the first derivative and i used the product rule and the chain rule
g'(x)=(1-2x)^2 (1) + (2(-2)(1-2x)
=(1-2x)^2 + -4(1-2x)
=2x^2 +8x -3

3. Originally Posted by surffan
g(x)=(1-2x)^2(x-3)
I need to state whether the function is odd or even and i said odd since there is a degree of 3, and the x and y intercepts i solved for and ended up with y=-2 and x=3,1/2. now i need to find the first and second derivative, and i was wondering if i should use the chain rule and product rule ?
It is even if and only if g(x)=g(-x) and odd if and only if g(x)=-g(-x).

So does g(1)=g(-1)?

or does g(1)=-g(-1)?

CB

4. i'll solve it like that then, i didnt think of that, thats what you get for working on these problems at 2am. do my derivatives look right though?

5. Originally Posted by surffan
i'll solve it like that then, i didnt think of that, thats what you get for working on these problems at 2am. do my derivatives look right though?

CB

6. how should i approach this function in order to get the derivatives? Since its factored im not really sure what i'm suppose to do.

7. Originally Posted by surffan
how should i approach this function in order to get the derivatives? Since its factored im not really sure what i'm suppose to do.
Product rule followed by chain rule:

$\frac{d}{dx}\left[(1-2x)^2(x-3)\right]=\left[\frac{d}{dx}(1-2x)^2 \right] (x-3) + (1-2x)^2\left[ \frac{d}{dx}(x-3) \right]$

CB