What Constitutes a Decreasing Interval? Problem:

For what values of $x$ is $f(x)=x^4-4x^3$ increasing? decreasing?

The first half of the answer in the book is:

if $x<3$, $f'(x)\leq 0$ and $f$ is decreasing

There is also a note:

Note that $f$ is decreasing at $x=0$ even though $f'(0)=0$

If $f'(0)=0$, then $f$ at $x=0$ is neither increasing nor decreasing. So why is the book saying what it's saying?
 A: To speak about decreasing or increasing, we do not need the derivative to even exist!
Recall that $f$ is called (strictly) decreasing on an interval $I$ if for all $x,y\in I$ with $x<y$, we have $f(x)>f(y)$.
We call $f$ decreasing at $x_0$ if it is decreasing on an open interval containing $x_0$.
If the derivative $f'(x_0)$ exists, then it is sufficient for $f$ to be decreasing at $x_0$ that $f'(x_0)<0$ and it is necessary that $f'(x_0)\le 0$.
The case that $f'(x_0)=0$ needs special attention. In your specific example, the MVT may come in handy.
A: By definition, $f$ is strictly decreasing on an interval $I$ if for each $x_1, x_2$ in $I$ with $x_1 < x_2$, $f(x_1) > f(x_2)$.  
Consider the interval $(-1, 1)$ that contains $0$.  
Since $f(x) = x^4 - 4x^3$, $f(0) = 0$.  
If $-1 < x < 0$, then $f(x) = x^4 - 4x^3 = x^3(x - 4)$ is the product of two negative numbers, so it is positive.  Hence, $f(x) > f(0)$, which means the function decreases between $x$ and $0$.
If $0 < x < 1$, then $f(x) = x^4 - 4x^3 = x^3(x - 4)$ is the product of a positive number and a negative number, so it is negative.  Hence, $f(0) > f(x)$, which means the function decreases between $0$ and $x$.
Putting the two observations together, we see that $f$ is decreasing at $x = 0$, as shown in the graph below.

A: Because the derivative of $f$ on both sides of $0$ is decreasing, the function is also decreasing at that point.
If the signs of the derivatives differed, it would be a maximum/minimum.
