Calculus: Isn't the velocity & tangent at a critical point = 0, and therefore a function is not increasing at that point? I encountered a question from Khan Acad that asked, at what intervals of x does the function increase. My intuition is that all ranges except for 0 and 1, because tangents at those points are flat per green graph below.
Backed up by the red graph, showing h'(1)=0 & h'(0)=0, zero velocity.
I need help figuring out my knowledge gap:

*

*Am I misinterpreting what "increasing" here?

*Am I missing the point of what a critical point is..?




 A: Increasing functions and strictly increasing functions are both well-defined terms with different definitions.
From wikipedia:

In calculus, a function $f$ defined on a subset of the real numbers
with real values is called monotonic if and only if it is either
entirely non-increasing, or entirely non-decreasing. That is, a
function that increases monotonically does not exclusively have to
increase, it simply must not decrease.
A function is called monotonically increasing (also increasing or
non-decreasing), if for all $x$ and $y$ such that $x\leq y$ one has
$f(x)\leq f(y),$ so $f$ preserves the order. Likewise, a function is called
monotonically decreasing (also decreasing or non-increasing) if, whenever
$x\leq y$, then $f(x)\geq f(y)$, so it reverses the order.
If the order $\leq$ in the definition of monotonicity is replaced by
the strict order $<$, then one obtains a stronger requirement. A
function with this property is called strictly increasing. Again, by
inverting the order symbol, one finds a corresponding concept called
strictly decreasing. A function may be called strictly monotone if it
is either strictly increasing or strictly decreasing. Functions that
are strictly monotone are one-to-one (because for $x$ not equal to
$y$, either $x<y$ or $x>y$ and so, by monotonicity, either $f(x) <$
$f(y)$ or $f(x) > f(y)$ (thus $f(x) \neq f(y)$.)

As B.Martin pointed out in the comments, $f(x)=x^3$ is a strictly increasing function (!), with $f′(0)=0$. Also, the constant function $f(x)=1$ is an increasing (but not strictly increasing) function with a derivative that is $0$ everywhere. So one always has to be clear what one is talking/asking about: whether or not a function is increasing or strictly increasing? At a point in the domain, or a subset of the domain, or over the entire domain?
Your/Khan academy's question is: Is the function $f(x) = 6x^5 - 15x^4 + 10x^3$ increasing (but not necessarily strictly increasing) for all $x \in \mathbb{R}$?
The answer is yes. In fact, your function is strictly increasing for all $x \in \mathbb{R}$, exactly the same as how $f(x) = x^3$ is strictly increasing for all $x \in \mathbb{R}$.
We may also talk about a function being strictly increasing at a point. See: https://math.stackexchange.com/a/364619/29156 for details. (Although, Did's definition of increasing at a point is really a definition of "strictly increasing at a point", since we are going with wikipedia's definitions).
The domain $D$ of all points for which $f(x)$ is increasing is the set $D = $ { $x: f(x)$ is increasing at the point $x$ }.
The function $f(x) = 6x^5 - 15x^4 + 10x^3$ is in fact strictly increasing at the point $x=0$, and it's also true that $f'(0)=0$. However, we only need to show that this function is increasing, and we can prove this using the definition of "increasing", or we can use the fact that $f'(x) \geq 0$ $\forall x \in \mathbb{R}$, which can be seen from the graph, or by noting that $f'(x) = (\sqrt{30}x(x-1))^2 \geq 0 \ \forall x \in \mathbb{R}$.
Going back to your question, "... because tangents at those points are flat..."
What is your definition of "flat"?
Useful:
A (differentiable) function $f$ is increasing but not strictly increasing at a point $x=c$, i.e. $f'(c) = 0$, $\iff \exists \epsilon > 0 $ such that $f$ is constant on the interval $ (c - \epsilon, c + \epsilon)$. I think this fact also works for all functions, not just differentiable or continuous ones.
A: A function $f$ is increasing if $f(b) ≥ f(a)$ with $b ≥ a$, for all $a,b$ in the domain, which is $x \in \mathbb R$ in your case. Note the $≥$ sign instead of $>$.
What you are thinking of is a strictly increasing function, otherwise called a monotonically increasing function. These functions can have (* see edit) $f'(x) = 0$, whereas an increasing function can $f'(x) = 0$ for a finite number of $x$ values.
Note that the derivative does not need to be defined at all points for a function to be an increasing function. For example, the function $g(x) = x^{1/3}$ increases for all $x \in \mathbb R$, despite the derivative not being defined at $x = 0$:

It is also possible for a function to be increasing in a restricted domain such as $(a,b)$, with other variations depending on whether the interval is open or closed.

Edit: Strictly increasing functions can have $f'(x) = 0$, as long as $f(x)$ is increasing in the neighbourhood of that point, say $(x - \epsilon, x + \epsilon)$ as mentioned by Adam Rubinson.
