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How can I find the critical/turning points of the implicit function?

Critical point is a wide term and is used in many branches of mathematics, but, the notion is in most cases connected to the derivative of a function or mapping.

• For real functions, a critical point is a point in the domain of the function where the function is either not differentiable or the derivative is equal to zero.
• Likewise, for a function of several real variables, a critical point is a value in its domain where the gradient is undefined or equals to zero.
• For complex variable functions, a critical point is, similarly, a point in the function's domain where it is either not holomorphic or the derivative is equal to zero.

The value of the function at a critical point (or equivalently a stationary point) is called a critical value.

If \(C \) is a plane curve, defined by an implicit function \[ f(x, y) = 0, \] the critical points of the projection onto the \(x\)-axis, parallel to the \(y\)-axis are the points where the tangent to \(C\) are parallel to the \(y\)-axis, that is the points where
\begin{align}
\frac{\partial f(x,y)}{\partial y} = 0.
\end{align}
In other words, the critical points are those points where the implicit function theorem does not apply. Find out more about critical points, including their extension to differentiable maps at Wikipedia.

A critical point should be differentiated from a stationary point. A critical point is all-encompassing, and occurs when \(f' = 0\) or when \(f \) is not differentiable, while a stationary point occurs when \(f' = 0\). Furthermore, points, where \( f' \) is not defined, are called singular points. With this interpretation in mind, we see that all stationary points (which may be minima, maxima, or inflection (turning) points) are critical points but not all critical points are stationary points.

Let's now consider a classical problem.

Find coordinates of all stationary points (all stationary points are critical points) for the curve
\begin{align}\label{eq1}
x^2 - 4xy - y^2 + 20 = 0.
\end{align}
As a bonus explanation, this curve is a conic section which takes the general form
\begin{align}
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0,
\end{align}
where \( A = 1, \ B = -4, \ C = -1, \ D = 0, \ E = 0, \ F = 20\). I f we check the discriminant of the general formula for conic sections,
\begin{align}
B^2 - 4AC = (-4)^2 - 4(1)(-1) = 16 + 4 = 20 > 0,
\end{align}
implying that Equation (\ref{eq1}) represents a hyperbola. It also represents a rectangular hyperbola since \( A + C = 0\). Check its shape at

https://www.geogebra.org/m/WXb2Nymr
which looks like Now, let's find our stationary points.

We differentiate the implicit function (\ref{eq1}) with respect to \( x \) using product rule:
\begin{align}\label{eq2}
2x - \Big[4x \frac{{\rm d}y}{{\rm d}x} + 4y \Big] = 2y\frac{{\rm d}y}{{\rm d}x} \implies x - 2x \frac{{\rm d}y}{{\rm d}x} - 2y = \frac{{\rm d}y}{{\rm d}x}.
\end{align}
Rearranging (\ref{eq2}) gives
\begin{align}
\frac{{\rm d}y}{{\rm d}x} = \frac{x - 2y}{2x + y}.
\end{align}
At stationary points, we must have \( {\rm d}y/{\rm d}x = 0\), and thus the equation above reduces to
\begin{align}
\frac{x - 2y}{2x + y} = 0.
\end{align}
This then tells us, the numerator of the equation above,
\begin{align}\label{eq3}
x - 2y = 0, \implies x = 2y.
\end{align}
Substituting Equation (\ref{eq3}) into Equation (\ref{eq1}), we have
\begin{align}
\begin{split}
(2y)^2 - 4(2y)y - y^{2} + 20 &\quad = 0\\
5y^2 &\quad = 20\\
y^2 &\quad = 4\\
y &\quad = \pm 2.
\end{split}
\end{align}
Substituting \( y = \pm 2 \) back into Equation (\ref{eq3}), we get \( x = 4 \) for \( y = 2 \) and \( x = -4 \) when \( y = -2 \).

Thus, the stationary points are \( (4, 2)\) and \( (-4, -2) \).

Can you classify these critical/stationary points?

For basics in MathJax and LaTeX typesetting, check the tutorial here.

Well, now the question is how to classify them, can we distinguish whether they are minimum, maximum or point of inflections?

Find the maximum and minimum values of \(y\) given by the equation
\begin{align}
x^3 + y^3 - 3xy = 0.
\end{align}

From the previous solution in post #2, to find the nature of stationary points, start from
\begin{align}
\frac{{\rm d}y}{{\rm d}x} = \frac{x - 2y}{y + 2x}.
\end{align}
We need to check the second derivative of the above equation at each stationary point. Thus, rearranging the equation above and differentiating with respect to \( x\) using product rule gives
\begin{align}
(y+ 2x) \frac{{\rm d}y}{{\rm d}x} = x - 2y,
\end{align}
\begin{align}\label{eq4}
(y + 2x) \frac{{\rm d}^{2}y}{{\rm d}x^{2}} + \Bigg(\frac{{\rm d}y}{{\rm d}x} + 2\Bigg)\frac{{\rm d}y}{{\rm d}x} = 1 - 2\frac{{\rm d}y}{{\rm d}x}.
\end{align}
We know that, at stationary point \( (4, 2)\), \( x = 4\), \( y = 2\), \( {\rm d}y/{\rm d}x = 0 \); and thus substituting these into Equation (\ref{eq4}) above yields
\begin{align}
\frac{{\rm d}^{2}y}{{\rm d}x^{2}} = \frac{1}{10} > 0.
\end{align}
Therefore, we conclude that the stationary point \((4, 2)\) is a local minimum.

In the like manner, we test the second stationary point \((-4, -2)\), by substituting \( x=-4\), \(y = -2\), \({\rm d}y/{\rm d}x = 0\) into (\ref{eq4}), where we get
\begin{align}
\frac{{\rm d}^{2}y}{{\rm d}x^{2}} = -\frac{1}{10} < 0,
\end{align}
and thus the stationary point \((-4, -2)\) is a local maximum.

Try to solve the other equation.

Thank you sir, well understood, and noted.

Once more from Equation (\ref{eq1}) in post #2,
\begin{align}\label{eq5}
x^2 - 4xy - y^2 + 20 = 0,
\end{align}

it is always important to specify how variables vary with the function/equation. For example, if we explicitly write
\begin{align}\label{eq6}
f(x) = x^2 - 4xy - y^2 + 20,
\end{align}
this means \(f\) is solely a function of \(x\) and in this sense \(y\) is constant. We would then proceed as follows:

\begin{align}
\begin{split}
\frac{{\rm d}f(x)}{{\rm d}x} &\quad = \frac{{\rm d}}{{\rm d}x}\Big[ x^2\Big] - \frac{{\rm d}}{{\rm d}x}\Big[ 4xy \Big] - \frac{{\rm d}}{{\rm d}x}\Big[ y^2\Big]\\
&\quad = \frac{{\rm d}}{{\rm d}x}\Big[ x^2\Big] - 4y\frac{{\rm d}}{{\rm d}x}\Big[ x \Big] - \frac{{\rm d}}{{\rm d}x}\Big[ y^2\Big]\\
&\quad = 2x - 4y.
\end{split}
\end{align}
To find the critical points of the function, we set the derivative equal to \(0\),
\begin{align}
\frac{{\rm d}f(x)}{{\rm d}x} = 2x - 4y = 0,
\end{align}
and then perform the second derivative test to determine their nature.

Terminologies

Terminologies used in this type of calculus are importatnt to note:

The maxima and minima (the respective plurals of maximum and minimum) of a function, are collectively known as extrema (the plural of extremum). These are the largest and smallest value of the function, either within a given range (the local or relative extrema) or on the entire domain of interest of a function (the global or absolute extrema).

Therefore, the global maximum is the largest value attained by a function in a given domain, while local maximum is the largest value attained by a function in some neighbourhood. Mathematically,

Consider any two real numbers \(a\), \(b\), such that \(a<b\); and let the real-valued function of these real variables be defined as
\begin{align}
f:[a,b] \rightarrow \mathcal{R}.
\end{align}
Then:

1. The function \(f\) is said to have an absolute/global maximum at \(a \le x_{0} \le b\) if and only if \(f(x_{0}) \ge f(x)\) for all \(x \in [a,b]\). The global/absolute minimum of a function can be defined in the like manner.
2. The function \(f\) is said to have a local or relative maximum at \(a \le x_{0} \le b\) if and only if there exists an open interval \(I\) (some range within the considered domain) about \(x_{0}\) such that \(f(x_{0}) \ge f(x)\) for all \(x\in I\); that's, if there exists epsilon, \(\epsilon >0\), such that \(f(x_{0}) \ge f(x)\) for all \(x \in (x_{0} - \epsilon, x_{0} + \epsilon)\). A local minimum of a function can be similarly defined.

As we have previously mentioned, if the function \(f\) is differentiable, then at the point of local extrema (local minimum or local maximum), the derivative of the function is zero. However, there exists functions, which are non differentiable, but still have local extrema (extreme values).

We illustrate these concepts graphically by assuming some function whose graph is shown below.

If the function is differentiable, and there is a local maxima/minima, then it is guaranteed that the derivative, at that point must be zero.

Local maxima and minima denoted by green bars.

One important point to note is that, a local maximum value of a function can be smaller than a local minimum value.

A critical point may be a local maximum/minimum/ or a saddle. A saddle is point on the surface of the graph of a function where the derivative in orthogonal directions are all zero, but which is not a local extremum (not local maximum nor local minimum) of the function.

A saddle (red point) on the graph of \(z=x^2−y^2\) (hyperbolic paraboloid), see more here.

For a continuous function \(f(x,y)\) on some domain, finding extrema and classifying their nature, may proceed as follows:

• Find the interior critical points.
• For each side of the boundary including vertices, find the extrema for that side.
• Among all the interior critical points and extrema on the boundary, find the maximum and minimum.

Let's illustrate how to find a local maximum/minimum or a saddle of a two variable function \(f(x,y)\). The steps would be as follows:

1. Find the input points \((x_{0},y_{0})\) where the gradient of \(f\) is zero -- the points where the tangent plane on the graph of \(f\) is flat:
   \[ \nabla f(x_{0}, y_{0}) = 0.\]
2. Perform the second partial derivative test. This step will help us to identify whether the point \((x_{0},y_{0})\) is a local maximum/minimum or a saddle. This step can further be carried as follows:
   • Compute the determinant of the Hessian matrix
     \begin{align}
     |Hf(x,y)| = {\rm det} \Bigg(\begin{bmatrix}
     f_{xx} & f_{yx} \\
     f_{xy} & f_{yy}
     \end{bmatrix} \Bigg)
     = f_{xx}(x_{0},y_{0}) f_{yy}(x_{0},y_{0}) - f_{xy}(x_{0},y_{0})^{2} \ \ \ \ \ (\text{since} \ \ f_{xy} = f_{yx}).
     \end{align}
   • If \(H<0\), then \((x_{0},y_{0})\) is a saddle point.
   • If \(H>0\), then \((x_{0},y_{0})\) is either a maximum or a minimum point. Thus, we need further tests to identify the true nature of this point:
     • If \( f_{xx}(x_{0},y_{0}) < 0 \) or \( f_{yy}(x_{0},y_{0}) < 0 \), then \( (x_{0},y_{0}) \) is a local maximum.
     • If \( f_{xx}(x_{0},y_{0}) > 0 \) or \( f_{yy}(x_{0},y_{0}) > 0 \), then \( (x_{0},y_{0}) \) is a local minimum.
   • If \(H = 0\), we do not have sufficient information to tell.

To find the global/absolute maximum/minimum, we have to rule out all possibilities in the given domain of interest.

With this information at hand, it is now easy to find the critical points and their nature, if we write the equation in post #4 as a two-variable function \(\ {\color {green} f(x,y) = x^3 + y^3 - 3xy}\).

Well understood, thank you.