1. ## First Derivative Test

I have to get the following:
- relative extrema of f
- values of f at which the relative extrema occurs
- intervals on which f is increasing
- intervals on which f is decreasing

when f(x) = $(1-x)^2 (1+x)^3$

Now when get to have the first derivative by multiplication rule f'(x) = g(x)*h'(x)+h(x)*g'(x):
f'(x) = $((1-x)^2)(3(1+x)^2)+((1+x)^3)(2(1-x))$
is it correct to say that f'(x)=0 when x=1 or x=-1?

and if it is, by substituting 1 and -1 to f(x), i'll arrive on ordered pairs' (1,0),(-1,0) which are on a vertical line. when i checked if the interval -1 < x < 1 is increasing or decreasing, i arrived at an answer that it is increasing which is not possible considering the locations of the two critical points.

Where did I go wrong?

2. When deriving $y=\; \left( 1-x \right)^{2}\left( 1+x \right)^{3}$ you have to keep in mind that the derivative of (1-x)^2 is -2(1-x) which is negative and instead of
$((1-x)^2)(3(1+x)^2)+((1+x)^3)(2(1-x))$
you should get this:
$((1-x)^2)(3(1+x)^2)-((1+x)^3)(2(1-x))$

Also, you are correct that -1 and 1 are both zeros but you can't just assume those are the only ones.

$((1-x)^2)(3(1+x)^2)=((1+x)^3)(2(1-x))$ which simplifies to:
$3 x^4-6 x^2+3 = -2 x^4-4 x^3+4 x+2$

and eventually simplifies to reveal a third zero of x = $1/5$.

You should find that the function is increasing from x = {-infinity,1/5},{1,infinity} and decreasing from x = {1/5,1} As shown in the graph of f.