You want to find the important characteristics of your function $f(x)$.
These include, but are not limited to:
- The domain of the function (if could be specified explicitly, or it may be the "natural domain", which you need to determine from the formula).
- The interecepts with the axes ($x$- and $y$-intercepts).
- The local extremes.
- The intervals where the function is increasing, decreasing, or constant.
- The intervals where the function is concave up, and where it is concave down.
- The global extremes, if any.
- The end behaviour, if any.
A function $f(x)$ can only have local extremes at points where (i) it is defined; and (ii) has a critical point. However, not every critical point needs to be a local extreme. A critical point is a point where the function is defined, and either the derivative does not exist or the derivative is equal to $0$.
If the derivative is continuous, then the only places where the derivative could change sign is at the critical points: this follows from the intermediate value theorem. If the derivative is positive at $a$ and negative at $b$, and is continuous on $[a,b]$, then it has to go through zero somewhere in between.
That means that if $p_1$ and $p_2$ are critical points of $f(x)$, and there is no critical points between $p_1$ and $p_2$, then the derivative has to be either always positive on $(p_1,p_2)$, or always negative on $(p_1,p_2)$. There are two ways to determine which it is: you can do a "sign analysis" of the derivative, which can be complicated. Or you can just plug in any number in $(p_1,p_2)$ into the derivative (any number, whichever one is simplest), and see if the derivative is positive there or negative there. That will tell you what sign the derivative is on the entire interval, because you know it has only one sign in that entire interval.
Once you know the critical points, you can pick any point between two of them to determine if the derivative is positive or negative on that interval.
For example, if
$$f(x) = \frac{x}{1+x^2},$$
then $$f'(x) = \frac{1-x^2}{(1+x^2)^2}.$$
The critical points are $x=1$ and $x=-1$. That means, since the derivative is continuous, that the only places where $f'(x)$ can change from positive to negative (the only places where $f(x)$ can change from increasing to decreasing), and the only places where $f(x)$ can change from negative to positive (the only places where $f(x)$ can change from decreasing to increasing) are the points where $f'(x)=0$, i.e., $x=-1$ and $x=1$. That means that the derivative has to be either always positive or always negative on $(-\infty,-1)$; always positive or always negative on $(-1,1)$; and always positive or always negative on $(1,\infty)$.
How do you tell which? You can do a sign analysis, or you can just plug in any number! Plug in any number in $(-\infty,-1)$ into $f'(x)$: if you get a negative answer, then $f'(x)$ has to be "always negative" on $(-\infty,-1)$. If you get a positive answer, then $f'(x)$ has to be "always positive" on $(-\infty,-1)$. Which number do you plug in? Any number; whatever number makes it easy to figure out. For instance, plug in $x=-2$; you get $f'(-2) = \frac{-3}{25}\lt 0$, so $f'(x)$ is always negative on $(-\infty,-1)$.
How do you tell whether $f'(x)$ is always positive or always negative on $(-1,1)$? You can plug in any number that is in $(-1,1)$ and that will tell you: for instance, $x=0$ is easy, and $f'(0) = 1\gt 0$, so $f'(x)$ is always positive on $(-1,1)$.
How do you tell whether $f'(x)$ is always positive or always negative on $(1,\infty)$? You can plug in any number that is in $(1,\infty)$ and see if you get a positive number of a negative number; that will tell you which sign $f'(x)$ is on the entire interval because $f'(x)$ cannot change signs there (because there are no critical points; it can't go from positive to negative or from negative to positive without either a break, or going through the $x$-axis).
Once you have that, you can use it to determine that $f'(x)$ is negative on $(-\infty,-1)$, positive on $(-1,1)$, and negative on $(1,\infty)$, which in turn tells you that $f(x) $is decreasing on $(-\infty,-1]$, increasing on $[-1,1]$, and decreasing on $[1,\infty)$, which gives you the local extremes (which are points where $f(x)$ changes from increasing to decreasing or from decreasing to increasing).
Points of inflection work the same way, because:
- $f(x)$ is concave up exactly where $f'(x)$ is increasing;
- $f(x)$ is concave down exactly where $f'(x)$ is decreasing;
- $f(x)$ changes concavity exactly where $f'(x)$ has local extremes (which are where $f'(x)$ changes from increasing to decreasing or from decreasing to increasing).
That means that the exact same logic that applies to finding critical points for $f(x)$ and figuring out where $f'(x)$ is positive and negative in order to find the local extremes of $f(x)$ can be used to find the points of inflection of $f(x)$: just apply that logic to $f'(x)$ and $f''(x)$ instead of to $f(x)$ and $f'(x)$, because you are just looking for the local extremes of $f'(x)$ (instead of $f(x)$), and that will give you the points of inflection of $f(x)$.
The First Derivative Test works because it tells you that $f(x)$ is changing from increasing to decreasing (if $f'(x)$ is switching from positive to negative at $p$), which means the function $f(x)$ is "going up" to $f(p)$, and then "coming down" from $f(p)$, which means $f(p)$ is a local maximum; or that $f(x)$ is switching from decreasing to increasing (if $f'(x)$ is switching from positive to negative at $p$), which means $f(x)$ is "going down" to $f(p)$ and then "going up" from $f(p)$, so $f(p)$ is a local minimum. Or that $f'(x)$ does not change signs, which means $f(x)$ is either "going up" to $f(p)$ and then keeps going up (so $f(p)$ is not a local extreme), or "goes down" to $f(p)$ and then keeps going down (so $f(p)$ is not a local extreme).
The Second Derivative Test works because if $f''(p)\gt 0$ that means $f'(x)$ is increasing around $p$. Since $f'(p)=0$ and $f'(x)$ is increasing, it has to be negative to the left of $p$ and positive to the right; in terms of the First Derivative Test, this tells you that $f(x)$ has a local minimum at $p$. And if $f''(p)\lt 0$, that means $f'(x)$ is decreasing near $p$, and since $f'(p)=0$, to be decreasing it has to be positive to the left of $p$ and negative to the right of $p$, which means (in terms of the First Derivative Test) that $f(x)$ has a local maximum at $p$.
If the shape is located and oriented in a "good" way relative to the $x$- and $y$-axes, the answer is yes, you can derive the second moment around any axis in the plane given only $I_x$ and $I_y.$
More generally, for an arbitrary shape at an arbitrary location and orientation, you will also need the product moment of inertia,
$$I_{xy} = -\iint_A xy \, dx dy.$$
It's also conventional in this context to write $I_{xx}$ and $I_{yy}$ instead of
$I_x$ and $I_y.$
Using this notation, then for a shape lying entirely in the $x,y$-plane
with the centroid of the shape at $(0,0),$
the moment of inertia around an axis given by a unit vector $\hat n$
can be computed (using matrix notation) by
the use of the inertia tensor, $I,$ as follows:
$$I_{\hat n} =
\hat n^T I \hat n =
\hat n^T
\begin{pmatrix}
I_{xx} & I_{xy} & 0 \\
I_{xy} & I_{yy} & 0 \\
0 & 0 & I_{xx} + I_{yy}
\end{pmatrix}
\hat n.
$$
Note that $\hat n$ can be, but is not required to be, a vector in the $x,y$-plane.
You can use the parallel axis theorem to get the moment of inertia around
an axis parallel to $\hat n$ that does not pass through the shape's centroid.
This is a special case of a more general formula for the moment of inertia of a
three-dimensional object around the axis through the centroid of the object
in the direction of the vector $\hat n$:
$$I_{\hat n} = \hat n^T
\begin{pmatrix}
I_{xx} & I_{xy} & I_{xz} \\
I_{xy} & I_{yy} & I_{yz} \\
I_{xz} & I_{yz} & I_{zz}
\end{pmatrix}
\hat n.
$$
(These notes offer this notation;
also see this page and the page that follows it,
or these lecture notes and the notes for the following lecture.)
For an object entirely in the $x,y$-plane, the $z$-coordinate of any point is zero,
hence $I_{xz} = I_{yz} = 0,$ and we can use the perpendicular axis theorem to
substitute for $I_{zz}.$
You can simplify the calculation further, however, by finding the principal axes
of your object. This is a set of orthogonal axes such that when the axis of rotation
is described by a vector $\hat n$ relative to those axes, the moment of inertia is simply
$$I_{\hat n} = \hat n^T
\begin{pmatrix}
I_{xx} & 0 & 0 \\
0 & I_{yy} & 0 \\
0 & 0 & I_{zz}
\end{pmatrix}
\hat n
$$
when the principal axes are taken as the $x$-, $y$-, and $z$-axes.
That is, if you place your object on your coordinate plane so that its principal
axes coincide with your $x$- and $y$-axes, or conversely choose $x$- and $y$-axes
that correspond with the object's principal axes, then you can derive the
moment of inertia around any axis merely by knowing its moments of inertia
around the two axes. But that assumes you are able to find the principal axes
in the first place, which (for an irregular body) may involve computing $I_{xy}$
in some initially chosen coordinate system
and then finding the eigenvectors of the inertia tensor in that coordinate system,
which indicate the principal axes.
A possibly useful fact: if your planar object has any bilateral symmetry,
the axis of symmetry is one of your principal axes.
That's easy to confirm by setting one of your coordinate axes to the axis
of symmetry and computing $I_{xy}$ in that coordinate system.
Best Answer
It makes sense to define $r$-th moment of a region $D \subseteq \mathbb{R}^n$ as $r$-th moment of the homogeneous solid body, occupying that region.
Thus, the $r$-th moment is defined as a rank $r$-tensor: $$ I^{(r)}_{i_1,i_2,\ldots,i_r} = \frac{\int_D \left(x_{i_1} x_{i_2}\cdots x_{i_r}\right) \mathrm{d}V}{\int_D \mathrm{d}V} $$