Calculus – Examining Critical Points in Similar Functions

analysiscalculusmultivariable-calculus

I am learning how to find extreme values of functions of two variables. I am getting familiar with critical point concept and I quite don't understand it. I believe I understand what stationary points are, though.

Why one function has no critical points and the other has $P(0,0)$ as its critical point?

Not sure if that's useful:

both functions have no local/global extremas
I believe both functions' domains and their derivatives' domains are all equal to each other, and it is $D = (\mathbb{R} \setminus\{ 0\}) \times \mathbb{R}$

Can someone help me understand why one function has critical point and the other does not? How do I find it?

Best Answer

For the function $f(x,y) = \frac{1}{x} + y$, the partial derivative with respect to $y$, $f_y(x,y) = 1$. This will never equal to $0$ or undefined. Therefore you can't find the critical point. Whereas, for the function $f(x,y) = \frac{1}{x} + y^2$, the partial derivative with respect to $y$, $f_y(x,y) = 2y$ which is equal to $0$ when $y=0$.

Related Solutions

[Math] Unique critical point does not imply global maximum/global minimum

Your intuitive picture is probably as follows: Imaging filling water into the hollow at the origin. As the water rises, since there are points lower than the origin, you expect the water to start spilling over into the lower lying areas at some point. And that point would need to be a critical point! The way to resolve the paradox is to note that before any spilling over occurs, the lake will in fact extend off to infinity. Look at a cross section of your function graph for fixed $y$, as $y$ starts at $0$ and then grows. In other words, study the function $x\mapsto x^2+y^2(1-x)^3$ as $y^2$ grows. You will see a local minimum, starting at the origin and approaching $x=1$ from below, and a local maximum, starting at infinity and approaching $x=1$ from above. As $y^2$ grows, the value at the local minimum grows, while the value at the local maximum decreases. So the aforementioned lake will extend out to $(1,\pm\infty)$ and then start to spill over the edge (the local maximum) at infinity before it spills over anywhere else.

A picture is tremendously helpful. If you have access to maple, try this:

plots[animate](plot,
  [x^2+y^2*(1-x)^3, x = 0 .. 2, view = [0 .. 2, -1 .. 2]], y = 0 .. 10);

Or you can take a look at this static plot of $f(x,y)$. It is easy to see how $(0,0)$ is a local, but not a global, maximum:

x^2+y^2(1-x)^3

[Math] Only one critical point, local minimum but not global

Using your example, one can show that similar examples exists for all $n\geq 3$: Let $F :\mathbb R^{2+n} \to \mathbb R$, $n\geq 1$ be defined by

$$F(x, y, z_1, \cdots, z_n) = f(x, y) + \epsilon(z_1^2 + \cdots + z_n^2)$$

where $f$ is the first example you gave. Then $dF=0$ only at the point $(0,1,0,\cdots, 0)$ and it is a local minimum by the second derivative test. Also $F(0,-3, 0,\cdots, 0) <F(0,-1, 0\cdots, 0)$ for some small $\epsilon$.

Best Answer

Related Solutions

[Math] Unique critical point does not imply global maximum/global minimum

[Math] Only one critical point, local minimum but not global

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