Where did i go wrong in trying to find the intervals where y is increasing and decreasing? Question: Find the intervals in which the following function is strictly increasing or decreasing: $(x+1)^3(x-3)^3$
The following was my differentiation:
$y = (x+1)^3(x-3)^3$
$\frac1y \frac{dy}{dx} = \frac3{x+1} + \frac3{x-3}$ (Through logarithmic differentiation)
This equation can be zero when x is ${-1, 1, 3}$ which gives us the intervals $(-\infty,-1),(-1,1),(1,3),(3,\infty)$
I checked that the last interval has a positive slope. Hence the second last should have a negative one, the third last a positive, and the first negative. However the book claims that the function is strictly decreasing in $(1,3),(3,\infty)$ and strictly decreasing in $(-\infty,-1), (-1,1)$
However, that doesn't make sense? If the function is increasing in both $(1,3)$ and $(3, \infty)$, why would it's slope be zero at 3? The function is obviously continuous.

$$\dfrac{dy}{dx}=3\{(x+1)(x-3)\}^2(x+1+x-3)$$
Now for real $x,\{(x+1)(x-3)\}^2\ge0$
So, the sign of $x+1+x-3$ will dictate the sign of $\dfrac{dy}{dx}$
