I'm starting to study Multivariable Calculus. I was learning about maxima and minima of a function. I know that we can use the method of equating the gradient to be $0$ and then finding the critical points and plugging in the values in are original function. My question is– what if my function $f(x,y)$ is such, that it has, say, a point $a$ where the function is indeterminate. When I find my gradient and equate it to $0$, then I get a few points to be my critical points, one of which is also $a$, the point where the function is indeterminate. At $a$, my function does not satisfy the assumption of being continuous or differentiable. Does this mean I can just drop it and assume that my other points are my maxima or minima? Can I even apply this method to begin with, if my function is not continuous and differentiable only at finitely many points? And what about if my function is not along a line?
Maxima minima of not everywhere continuous functions
continuityderivativesmaxima-minimamultivariable-calculus
Related Solutions
In this case you want to write down your function in sections: $$f(x)=\left\{\begin{array}{cc}\frac{1}{(x-2)+(x-4)},&x\geq 4,\\ \frac{1}{(x-2)-(x-4)},& 2\leq x < 4,\\ \frac{1}{-(x-2)-(x-4)},& x<2\end{array}\right.$$
On each interval you can now do your standard method. After that you compare your so found extremas with each other and $f$ evaluated at the boundary of your intervals, i.e. $f(2)$ and $f(4)$.
Edit: Your example $f(x)=\frac{1}{|x-4|}$ for reaching maximum or mininum: Formally speaking this $f$ does not attain its maximum or minimum value. Let us discuss why. $f$ is defined as $f:\mathbb{R}\setminus\{4\}\rightarrow\mathbb{R}$ and maximum value in $z$ for this $f$ would be defined as $$\exists z\in\mathbb{R}\setminus\{4\} \mbox{ such that } \forall x\in\mathbb{R}\setminus\{4\} \mbox{ we have } f(z)\geq f(x)$$ Let us assume such a $z\in\mathbb{R}\setminus\{4\}$ exists. Since $$\lim_{x\rightarrow4}f(x)=\infty$$ we find an $\tilde{x}\in\mathbb{R}\setminus\{4\} $ with $f(z)<f(\tilde{x})$. The same is true in this case for the minimum.
Essentially this is the difference between a supremum and a maximum. A supremum does not need to be attained while a maximum has to be attained by definition. See also https://en.wikipedia.org/wiki/Infimum_and_supremum
There is no global minima. To check this take $y\to-\infty$ and $x=1$ and you get $f(x,y)\to-\infty$ but $(0,0)$ is a local minimum since for every $0<\epsilon<1$:$$\forall |x|<\epsilon ,|y|<\epsilon\qquad\qquad \epsilon^2(1+(1+\epsilon)^3)<2\epsilon^2<f(x,y)<\epsilon^2(1+(1+\epsilon)^3)<9\epsilon^2$$for a close look:
and for a further look:
Best Answer
Nope. You can't 'just drop it' because you might drop a valid solution.
The absolute value function $x\mapsto|x|$ is continuous but not differentiable at $x=0$, anyway it has a minimum there.
A modified step function $x\mapsto\begin{cases}0&x<0\\1&x\ge 0\end{cases}$ is semi-differentiable but not continuous at $x=0$, anyway it has a maximum there.
Similarly, a single-point characteristic function $x\mapsto\begin{cases}1&x=0\\0&\text{otherwise}\end{cases}$ is neither differentiable nor continuous at $x=0$ but it has a maximum there.