সন্ধি বিন্দু কী?

Here's some graphical intuition:

The top of a hill is always flat.

Good, "flat" at that point implies that the slope of the curve is 0 at that point.

The first derivative test tells us that at local extrema of a continuous function $f(x)$, the derivative of $f(x)$ is 0. The first derivative of a function is the slope of the graph at that point, and the slope at a maximum or minimum is 0 (as the graphic above demonstrates).

So to find local extrema given a function, you:

1. Derive the function

2. Set the derivative equal to zero and solve. The resulting x-values are the x-values of the extrema!

And to figure out whether the extremum you found is a minimum or a maximum:

1. Set up a number line, with the extremum/extrema as points on the line

2. Plug in values that are greater than and less than the extremum into the first derivative of the function

3. If values less than that extremum yield a positive first derivative and values greater than that extremum yield a negative first derivative, that extremum is a maximum. If we think about this intuitively, this makes sense! The slope of a graph leading up to a maximum is positive (there's an incline) and the slope of the graph after a maximum is negative (the graph declines)

4. The same is true for the opposite - so if values less than that extremum yield a negative first derivative and values greater than that extremum yield a positive first derivative, that extremum is a minimum. This also makes sense - the slope of a graph leading up to a minimum is negative (there's a decline) and the slope of the graph after a minimum is positive (the graph is at an incline)

Good! $f'(x) = -2x + 4$ so the (only) local extremum occurs at $x=2$.

(Videos) Finding Critical Numbers

by PatrickJMT

Here's a video by patrickJMT where he shows you how to find critical points.

A critical point is any point, that, when plugged into the derivative of the function, returns zero or makes it undefined. The critical point also has to be in the domain of $f(x)$. $$\boxed{\begin{align} &f'(c)=0\ \boxed{\text{OR}}\\ &f'(c)=\text{undefined}\\ \\ &\boxed{\text{AND}}\\ &c\text{ is in the domain of }f(x) \end{align}}$$

Let's start by finding the critical points of the function: $$f(x)=\frac{x^2}{x^2-9}$$

Take the derivative of the function. $$\begin{align} f'(x)&=\frac{\color{red}{[x^2]'}(x^2-9)-x^2\color{red}{[x^2-9]'}}{[x^2-9]^2}\\ &=\frac{2x(x^2-9)-x^2(2x)}{(x^2-9)^2}\\ &=\frac{2x[x^2-9-x^2]}{(x^2-9)^2}\\ &=\frac{2x[-9]}{(x^2-9)^2}\\ &=\frac{-18x}{(x^2-9)^2}\\ \end{align}$$

To find the critical point, we look at what values make this derivative equal zero or undefined. Let's start by finding where it equals zero. $$\begin{align} \frac{-18x}{(x^2-9)^2}&=0\\ -18x&=0\\ x&=0 \end{align}$$

But this derivative can also be undefined. This happens when the denominator is equal to zero. $$\begin{align} (x^2-9)^2&=0\\ x^2-9&=0\\ x^2&=9\\ x&=\pm3 \end{align}$$ Be careful though, if we plug either of these values back into the original function we get some number over zero. Since we cannot divide by zero, these values aren't in the domain of $f(x)$. So, $x=\pm3$ are NOT critical values.

Therefore, $x=0$ is our only critical value.

Here's another video where he looks for the critical points of another function.