Критичні точки та ознака екстремума черех першу похідну

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 $x=2$.

(Video) Finding Intervals of Increase, Decrease, Local Maximums, and Local Minimums

by PatrickJMT

Here's a video by patrickJMT where he shows you how to apply the first derivative test to a function to find the intervals on which it increases and decreases and where local minimums and maximums are.

He demonstrates this test with the function $f(x)=x^2$.

We can see on the graph that the function decreasing from negative infinity to zero. $$(-\infty,0)\ \searrow$$ The function then increases from zero to infinity. $$(0,\infty)\ \nearrow$$ Finally, there's a local minimum at $x=0$.

But we can also find all this information out by looking at the first derivative. This is helpful because sometimes we don't have the graph or don't want to draw one.

To apply the first derivative test we:

1. Take the derivative of the function.
2. Figure out where the derivative equals zero or is undefined. These are the critical points.
3. Look at the sign of the derivative. Plug in a value within each critical point's interval to see if the derivative is positive or negative. Positive means it's increasing. Negative means it's decreasing.
4. If it goes from decreasing to increasing, that critical point is a local min. If it goes from increasing to decreasing that critical point is a local max.

This might not make sense so let's try it with our quadratic function above. First, we take the derivative. $$\begin{align} f(x)&=x^2\\ f'(x)&=2x \end{align}$$

Next, set it equal to zero and solve. $$\begin{align} 2x&=0\\ x&=0 \end{align}$$

Since $x=0$ is within the domain of our original function we know that it is a critical point.

Now we take a look at the sign of the derivative. To do this draw a number line and place all critical points on the line. In this example, we only have the one.

We want to plug a number to the left of the critical point and to the right into the derivative in order to see if it's positive or negative. Let's try $x=-10$ and $x=2,000,000$. $$\begin{align} f'(-10)&=2\color{red}{(-10)}\\ &=-20\\ f'(2,000,000)&=2\color{red}{(2,000,000)}\\ &=4,000,000 \end{align}$$

Since $x=-10$ is to the left of our critical point and returns a negative value, all numbers to the left of $x=0$ give us a negative number. Similarly, since $x=2,000,000$ is to the right of our critical point and returns a positive value, all numbers to the right of $x=0$ give us a positive number.

Where ever our derivative is negative our original function is decreasing. Similarly, where ever our derivative is positive our original function is increasing. So, based on our sign chart, we can conclude that our original function is decreasing from negative infinity to zero and increasing from zero to infinity. $$(-\infty,0)\ \searrow\\ (0,\infty)\ \nearrow$$ Notice that this is exactly what we go above from looking at the graph!

Finally, we can look at the sign change around the critical points. Since it goes from decreasing (the left) to increasing (the right) this indicates that our critical point is a local minimum. And looking at the graph we can see that $x=0$ is a local minimum.

Similarly, if it goes from increasing to decreasing we know that the critical point is a local maximum.

Let's look at a harder example, in which we can't easily graph and see the answer. $$f(x)=(x^2-1)^3$$

The first step we do is take the derivative. $$\begin{align} f(x)&=(x^2-1)^3\\ f'(x)&=3(x^2-1)^2\cdot\color{red}{[x^2-1]}\\ &=3(x^2-1)^2\cdot2x\\ &=(6x)(x^2-1)^2 \end{align}$$

Now we look for our critical points: where the derivative equals zero or is undefined. $$(6x)(x^2-1)^2=0\\ \begin{alignat}{3} &6x=0\quad&&\text{or}\quad&&(x^2-1)^2=0\\ &x=0\quad&&\text{or}\quad&&x^2=1\\ &\ &&\ &&x=\pm1 \end{alignat}$$

In this example, we have three critical numbers: $x=-1$, $x=0$, $x=1$. Let's go ahead and make a sign table with these.

We have four intervals we have to test: $(-\infty,-1)$, $(-1,0)$, $(0,1)$, and $(1,\infty)$. To test these, we need to choose a number within each interval. Let's begin with the interval greater than one and choose the number 2. If we plug any number into the square term, $(x^2-1)^2$, we will always get out a positive. And if we plug a positive into the $6x$ term we will always get a positive. A positive times a positive is always a positive.

The same logic can be applied to the intverval $(0,1)$ because all the numbers in the interval are positive.

For the two intervals less than zero, we will get a positive value in the square term and a negative value in the $6x$ term. A negative times a positive is always a negative.

As you can see, our original function is decreasing on the interval from negative infinity to zero and increasing on the interval zero to infinity. $$(-\infty,0)\ \searrow\\ (0,\infty)\ \nearrow$$

The only point where the derivative changes from increasing/decreasing to decreasing/increasing is at $x=0$. It is going from decreasing to increasing so we know that it is a local minimum.

To find the exact point of this local minimum we take the critical point, $x=0$, and plug it back into our original function. $$\begin{align} f(0)&=(\color{red}{(0)}^2-1)^3\\ &=(-1)^3\\ &=-1 \end{align}$$

The local minimum is at the point $(0,-1)$.

Finally, let's just take a quick look at the graph of this function to double check our work.

(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.