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.

• Consider the function $f(x)=x^3$ in the point $x=0$. The slope is $0$, but still, it's strictly increasing on both $(-\infty,0)$ and $(0,\infty)$. – vrugtehagel Mar 15 '16 at 15:00
• Think about points of inflection, which is also the case in the example of Vrugtehagel. Second. Please make a graph to see what's going on. Third, when solving $y'=0$, there are double zeros. What does that imply? – imranfat Mar 15 '16 at 15:02
• Just a question: why did you use logarithmic differentiation? There isn't any variable in the exponent. – Airdish Mar 15 '16 at 15:11
• @TheOddbodNumber I just learned about it so was practicing it. – Aayush Agrawal Mar 15 '16 at 15:50

$$\dfrac{dy}{dx}=3\{(x+1)(x-3)\}^2(x+1+x-3)$$
Now for real $x,\{(x+1)(x-3)\}^2\ge0$
So, the sign of $x+1+x-3$ will dictate the sign of $\dfrac{dy}{dx}$
• There is a tiny typo, i think you confused $x +1$ with $x + 2$. Also while i do understand that, that makes me even more confused, since then 2x - 2 is the determiner of slope sign, which would give only a single interval, even though the other method gives four intervals. – Aayush Agrawal Mar 15 '16 at 15:06
• @AayushAgrawal, try setting the values of $x$ near the intervals you have found – lab bhattacharjee Mar 15 '16 at 15:14