(Fig.1)
Figure 1. Derivative graph
Example 1
Using the graph (Fig. 2) of the derivative, determine at what point on the segment [-3; 5] function is maximum.
Figure 2. Derivative graph
Solution: On this segment, the derivative is negative, which means that the function decreases from left to right, and the largest value is on the left side at point -3.
Example 2
Using the graph (Fig. 3) of the derivative, determine the number of maximum points on the segment [-11; 3].
Figure 3. Derivative graph
Solution: The maximum points correspond to the points where the sign of the derivative changes from positive to negative. On this interval, the function changes sign from plus to minus twice - at point -10 and at point -1. This means the number of maximum points is two.
Example 3
Using the graph (Fig. 3) of the derivative, determine the number of minimum points in the segment [-11; -1].
Solution: The minimum points correspond to the points where the sign of the derivative changes from negative to positive. On this segment, such a point is only -7. This means that the number of minimum points on a given segment is one.
Example 4
Using the graph (Fig. 3) of the derivative, determine the number of extremum points.
Solution: The extreme points are both the minimum and maximum points. Let's find the number of points at which the derivative changes sign.
The derivative of a function $y = f(x)$ at a given point $x_0$ is the limit of the ratio of the increment of a function to the corresponding increment of its argument, provided that the latter tends to zero:
$f"(x_0)=(lim)↙(△x→0)(△f(x_0))/(△x)$
Differentiation is the operation of finding the derivative.
Table of derivatives of some elementary functions
| Function | Derivative |
| $c$ | $0$ |
| $x$ | $1$ |
| $x^n$ | $nx^(n-1)$ |
| $(1)/(x)$ | $-(1)/(x^2)$ |
| $√x$ | $(1)/(2√x)$ |
| $e^x$ | $e^x$ |
| $lnx$ | $(1)/(x)$ |
| $sinx$ | $cosx$ |
| $cosx$ | $-sinx$ |
| $tgx$ | $(1)/(cos^2x)$ |
| $ctgx$ | $-(1)/(sin^2x)$ |
1. The derivative of the sum (difference) is equal to the sum (difference) of the derivatives
$(f(x) ± g(x))"= f"(x)±g"(x)$
Find the derivative of the function $f(x)=3x^5-cosx+(1)/(x)$
The derivative of a sum (difference) is equal to the sum (difference) of derivatives.
$f"(x) = (3x^5)"-(cos x)" + ((1)/(x))" = 15x^4 + sinx - (1)/(x^2)$
2. Derivative of the product
$(f(x) g(x)"= f"(x) g(x)+ f(x) g(x)"$
Find the derivative $f(x)=4x cosx$
$f"(x)=(4x)"·cosx+4x·(cosx)"=4·cosx-4x·sinx$
3. Derivative of the quotient
$((f(x))/(g(x)))"=(f"(x) g(x)-f(x) g(x)")/(g^2(x)) $
Find the derivative $f(x)=(5x^5)/(e^x)$
$f"(x)=((5x^5)"·e^x-5x^5·(e^x)")/((e^x)^2)=(25x^4·e^x- 5x^5 e^x)/((e^x)^2)$
4. The derivative of a complex function is equal to the product of the derivative of the external function and the derivative of the internal function
$f(g(x))"=f"(g(x)) g"(x)$
$f"(x)=cos"(5x)·(5x)"=-sin(5x)·5= -5sin(5x)$
If a material point moves rectilinearly and its coordinate changes depending on time according to the law $x(t)$, then the instantaneous speed of this point is equal to the derivative of the function.
The point moves along the coordinate line according to the law $x(t)= 1.5t^2-3t + 7$, where $x(t)$ is the coordinate at time $t$. At what point in time will the speed of the point be equal to $12$?
1. Speed is the derivative of $x(t)$, so let’s find the derivative of the given function
$v(t) = x"(t) = 1.5 2t -3 = 3t -3$
2. To find at what point in time $t$ the speed was equal to $12$, we create and solve the equation:
Recall that the equation of a straight line that is not parallel to the coordinate axes can be written in the form $y = kx + b$, where $k$ is the slope of the straight line. The coefficient $k$ is equal to the tangent of the angle of inclination between the straight line and the positive direction of the $Ox$ axis.
The derivative of the function $f(x)$ at the point $х_0$ is equal to the slope $k$ of the tangent to the graph at this point:
Therefore, we can create a general equality:
$f"(x_0) = k = tanα$
In the figure, the tangent to the function $f(x)$ increases, therefore the coefficient $k > 0$. Since $k > 0$, then $f"(x_0) = tanα > 0$. The angle $α$ between the tangent and the positive direction $Ox$ is acute.
In the figure, the tangent to the function $f(x)$ decreases, therefore, the coefficient $k< 0$, следовательно, $f"(x_0) = tgα < 0$. Угол $α$ между касательной и положительным направлением оси $Ох$ тупой.
In the figure, the tangent to the function $f(x)$ is parallel to the $Ox$ axis, therefore, the coefficient $k = 0$, therefore, $f"(x_0) = tan α = 0$. The point $x_0$ at which $f "(x_0) = 0$, called extremum.
The figure shows a graph of the function $y=f(x)$ and a tangent to this graph drawn at the point with the abscissa $x_0$. Find the value of the derivative of the function $f(x)$ at point $x_0$.
The tangent to the graph increases, therefore, $f"(x_0) = tan α > 0$
In order to find $f"(x_0)$, we find the tangent of the angle of inclination between the tangent and the positive direction of the $Ox$ axis. To do this, we build the tangent to the triangle $ABC$.
Let's find the tangent of the angle $BAC$. (The tangent of an acute angle in a right triangle is the ratio of the opposite side to the adjacent side.)
$tg BAC = (BC)/(AC) = (3)/(12)= (1)/(4)=$0.25
$f"(x_0) = tg BAC = 0.25$
Answer: $0.25$
The derivative is also used to find the intervals of increasing and decreasing functions:
If $f"(x) > 0$ on an interval, then the function $f(x)$ is increasing on this interval.
If $f"(x)< 0$ на промежутке, то функция $f(x)$ убывает на этом промежутке.
The figure shows the graph of the function $y = f(x)$. Find among the points $х_1,х_2,х_3...х_7$ those points at which the derivative of the function is negative.
In response, write down the number of these points.
(Fig.1)
Figure 1. Derivative graph
Example 1
Using the graph (Fig. 2) of the derivative, determine at what point on the segment [-3; 5] function is maximum.
Figure 2. Derivative graph
Solution: On this segment, the derivative is negative, which means that the function decreases from left to right, and the largest value is on the left side at point -3.
Example 2
Using the graph (Fig. 3) of the derivative, determine the number of maximum points on the segment [-11; 3].
Figure 3. Derivative graph
Solution: The maximum points correspond to the points where the sign of the derivative changes from positive to negative. On this interval, the function changes sign from plus to minus twice - at point -10 and at point -1. This means the number of maximum points is two.
Example 3
Using the graph (Fig. 3) of the derivative, determine the number of minimum points in the segment [-11; -1].
Solution: The minimum points correspond to the points where the sign of the derivative changes from negative to positive. On this segment, such a point is only -7. This means that the number of minimum points on a given segment is one.
Example 4
Using the graph (Fig. 3) of the derivative, determine the number of extremum points.
Solution: The extreme points are both the minimum and maximum points. Let's find the number of points at which the derivative changes sign.
Let's imagine a straight road passing through a hilly area. That is, it goes up and down, but does not turn right or left. If the axis is directed horizontally along the road and vertically, then the road line will be very similar to the graph of some continuous function:
The axis is a certain level of zero altitude; in life we use sea level as it.
As we move forward along such a road, we also move up or down. We can also say: when the argument changes (movement along the abscissa axis), the value of the function changes (movement along the ordinate axis). Now let's think about how to determine the “steepness” of our road? What kind of value could this be? It’s very simple: how much the height will change when moving forward a certain distance. Indeed, on different sections of the road, moving forward (along the x-axis) by one kilometer, we will rise or fall by a different number of meters relative to sea level (along the y-axis).
Let’s denote progress (read “delta x”).
The Greek letter (delta) is commonly used as a prefix in mathematics, meaning "change". That is - this is a change in quantity, - a change; then what is it? That's right, a change in magnitude.
Important: an expression is a single whole, one variable. Never separate the “delta” from the “x” or any other letter! That is, for example, .
So, we have moved forward, horizontally, by. If we compare the line of the road with the graph of the function, then how do we denote the rise? Certainly, . That is, as we move forward, we rise higher.
The value is easy to calculate: if at the beginning we were at a height, and after moving we found ourselves at a height, then. If the end point is lower than the starting point, it will be negative - this means that we are not ascending, but descending.
Let's return to "steepness": this is a value that shows how much (steeply) the height increases when moving forward one unit of distance:
Let us assume that on some section of the road, when moving forward by a kilometer, the road rises up by a kilometer. Then the slope at this place is equal. And if the road, while moving forward by m, dropped by km? Then the slope is equal.
Now let's look at the top of a hill. If you take the beginning of the section half a kilometer before the summit, and the end half a kilometer after it, you can see that the height is almost the same.
That is, according to our logic, it turns out that the slope here is almost equal to zero, which is clearly not true. Just over a distance of kilometers a lot can change. It is necessary to consider smaller areas for a more adequate and accurate assessment of steepness. For example, if you measure the change in height as you move one meter, the result will be much more accurate. But even this accuracy may not be enough for us - after all, if there is a pole in the middle of the road, we can simply pass it. What distance should we choose then? Centimeter? Millimeter? Less is better!
In real life, measuring distances to the nearest millimeter is more than enough. But mathematicians always strive for perfection. Therefore, the concept was invented infinitesimal, that is, the absolute value is less than any number that we can name. For example, you say: one trillionth! How much less? And you divide this number by - and it will be even less. And so on. If we want to write that a quantity is infinitesimal, we write like this: (we read “x tends to zero”). It is very important to understand that this number is not zero! But very close to it. This means that you can divide by it.
The concept opposite to infinitesimal is infinitely large (). You've probably already come across it when you were working on inequalities: this number is modulo greater than any number you can think of. If you come up with the biggest number possible, just multiply it by two and you'll get an even bigger number. And infinity is even greater than what happens. In fact, the infinitely large and the infinitely small are the inverse of each other, that is, at, and vice versa: at.
Now let's get back to our road. The ideally calculated slope is the slope calculated for an infinitesimal segment of the path, that is:
I note that with an infinitesimal displacement, the change in height will also be infinitesimal. But let me remind you that infinitesimal does not mean equal to zero. If you divide infinitesimal numbers by each other, you can get a completely ordinary number, for example, . That is, one small value can be exactly times larger than another.
What is all this for? The road, the steepness... We’re not going on a car rally, but we’re teaching mathematics. And in mathematics everything is exactly the same, only called differently.
The derivative of a function is the ratio of the increment of the function to the increment of the argument for an infinitesimal increment of the argument.
Incrementally in mathematics they call change. The extent to which the argument () changes as it moves along the axis is called argument increment and is designated. How much the function (height) has changed when moving forward along the axis by a distance is called function increment and is designated.
So, the derivative of a function is the ratio to when. We denote the derivative with the same letter as the function, only with a prime on the top right: or simply. So, let's write the derivative formula using these notations:
As in the analogy with the road, here when the function increases, the derivative is positive, and when it decreases, it is negative.
Is it possible for the derivative to be equal to zero? Certainly. For example, if we are driving on a flat horizontal road, the steepness is zero. And it’s true, the height doesn’t change at all. So it is with the derivative: the derivative of a constant function (constant) is equal to zero:
since the increment of such a function is equal to zero for any.
Let's remember the hilltop example. It turned out that it was possible to arrange the ends of the segment on opposite sides of the vertex in such a way that the height at the ends turns out to be the same, that is, the segment is parallel to the axis:
But large segments are a sign of inaccurate measurement. We will raise our segment up parallel to itself, then its length will decrease.
Eventually, when we are infinitely close to the top, the length of the segment will become infinitesimal. But at the same time, it remained parallel to the axis, that is, the height difference at its ends is equal to zero (it does not tend to, but is equal to). So the derivative
This can be understood this way: when we stand at the very top, a small shift to the left or right changes our height negligibly.
There is also a purely algebraic explanation: to the left of the vertex the function increases, and to the right it decreases. As we found out earlier, when a function increases, the derivative is positive, and when it decreases, it is negative. But it changes smoothly, without jumps (since the road does not change its slope sharply anywhere). Therefore, there must be between negative and positive values. It will be where the function neither increases nor decreases - at the vertex point.
The same is true for the trough (the area where the function on the left decreases and on the right increases):
A little more about increments.
So we change the argument to magnitude. We change from what value? What has it (the argument) become now? We can choose any point, and now we will dance from it.
Consider a point with a coordinate. The value of the function in it is equal. Then we do the same increment: we increase the coordinate by. What is the argument now? Very easy: . What is the value of the function now? Where the argument goes, so does the function: . What about function increment? Nothing new: this is still the amount by which the function has changed:
Practice finding increments:
Solutions:
At different points with the same argument increment, the function increment will be different. This means that the derivative at each point is different (we discussed this at the very beginning - the steepness of the road is different at different points). Therefore, when we write a derivative, we must indicate at what point:
A power function is a function where the argument is to some degree (logical, right?).
Moreover - to any extent: .
The simplest case is when the exponent is:
Let's find its derivative at a point. Let's recall the definition of a derivative:
So the argument changes from to. What is the increment of the function?
Increment is this. But a function at any point is equal to its argument. That's why:
The derivative is equal to:
The derivative of is equal to:
b) Now consider the quadratic function (): .
Now let's remember that. This means that the value of the increment can be neglected, since it is infinitesimal, and therefore insignificant against the background of the other term:
So, we came up with another rule:
c) We continue the logical series: .
This expression can be simplified in different ways: open the first bracket using the formula for abbreviated multiplication of the cube of the sum, or factorize the entire expression using the difference of cubes formula. Try to do it yourself using any of the suggested methods.
So, I got the following:
And again let's remember that. This means that we can neglect all terms containing:
We get: .
d) Similar rules can be obtained for large powers:
e) It turns out that this rule can be generalized for a power function with an arbitrary exponent, not even an integer:
| (2) |
The rule can be formulated in the words: “the degree is brought forward as a coefficient, and then reduced by .”
We will prove this rule later (almost at the very end). Now let's look at a few examples. Find the derivative of the functions:
Here we will use one fact from higher mathematics:
With expression.
You will learn the proof in the first year of institute (and to get there, you need to pass the Unified State Exam well). Now I’ll just show it graphically:
We see that when the function does not exist - the point on the graph is cut out. But the closer to the value, the closer the function is to. This is what “aims.”
Additionally, you can check this rule using a calculator. Yes, yes, don’t be shy, take a calculator, we’re not at the Unified State Exam yet.
So, let's try: ;
Don't forget to switch your calculator to Radians mode!
etc. We see that the smaller, the closer the value of the ratio to.
a) Consider the function. As usual, let's find its increment:
Let's turn the difference of sines into a product. To do this, we use the formula (remember the topic “”): .
Now the derivative:
Let's make a replacement: . Then for infinitesimal it is also infinitesimal: . The expression for takes the form:
And now we remember that with the expression. And also, what if an infinitesimal quantity can be neglected in the sum (that is, at).
So, we get the following rule: the derivative of the sine is equal to the cosine:
These are basic (“tabular”) derivatives. Here they are in one list:
Later we will add a few more to them, but these are the most important, since they are used most often.
Practice:
Solutions:
There is a function in mathematics whose derivative for any value is equal to the value of the function itself at the same time. It is called “exponent”, and is an exponential function
The base of this function - a constant - is an infinite decimal fraction, that is, an irrational number (such as). It is called the “Euler number”, which is why it is denoted by a letter.
So, the rule:
Very easy to remember.
Well, let’s not go far, let’s immediately consider the inverse function. Which function is the inverse of the exponential function? Logarithm:
In our case, the base is the number:
Such a logarithm (that is, a logarithm with a base) is called “natural”, and we use a special notation for it: we write instead.
What is it equal to? Of course, .
The derivative of the natural logarithm is also very simple:
Examples:
Answers: The exponential and natural logarithm are uniquely simple functions from a derivative perspective. Exponential and logarithmic functions with any other base will have a different derivative, which we will analyze later, after we go through the rules of differentiation.
Rules of what? Again a new term, again?!...
Differentiation is the process of finding the derivative.
That's all. What else can you call this process in one word? Not derivative... Mathematicians call the differential the same increment of a function at. This term comes from the Latin differentia - difference. Here.
When deriving all these rules, we will use two functions, for example, and. We will also need formulas for their increments:
There are 5 rules in total.
If - some constant number (constant), then.
Obviously, this rule also works for the difference: .
Let's prove it. Let it be, or simpler.
Examples.
Find the derivatives of the functions:
Solutions:
Everything is similar here: let’s introduce a new function and find its increment:
Derivative:
Examples:
Solutions:
Now your knowledge is enough to learn how to find the derivative of any exponential function, and not just exponents (have you forgotten what that is yet?).
So, where is some number.
We already know the derivative of the function, so let's try to reduce our function to a new base:
To do this, we will use a simple rule: . Then:
Well, it worked. Now try to find the derivative, and don't forget that this function is complex.
Happened?
Here, check yourself:
The formula turned out to be very similar to the derivative of an exponent: as it was, it remains the same, only a factor appeared, which is just a number, but not a variable.
Examples:
Find the derivatives of the functions:
Answers:
It’s similar here: you already know the derivative of the natural logarithm:
Therefore, to find an arbitrary logarithm with a different base, for example:
We need to reduce this logarithm to the base. How do you change the base of a logarithm? I hope you remember this formula:
Only now we will write instead:
The denominator is simply a constant (a constant number, without a variable). The derivative is obtained very simply:
Derivatives of exponential and logarithmic functions are almost never found in the Unified State Exam, but it will not be superfluous to know them.
What is a "complex function"? No, this is not a logarithm, and not an arctangent. These functions can be difficult to understand (although if you find the logarithm difficult, read the topic “Logarithms” and you will be fine), but from a mathematical point of view, the word “complex” does not mean “difficult”.
Imagine a small conveyor belt: two people are sitting and doing some actions with some objects. For example, the first one wraps a chocolate bar in a wrapper, and the second one ties it with a ribbon. The result is a composite object: a chocolate bar wrapped and tied with a ribbon. To eat a chocolate bar, you need to do the reverse steps in reverse order.
Let's create a similar mathematical pipeline: first we will find the cosine of a number, and then square the resulting number. So, we are given a number (chocolate), I find its cosine (wrapper), and then you square what I got (tie it with a ribbon). What happened? Function. This is an example of a complex function: when, to find its value, we perform the first action directly with the variable, and then a second action with what resulted from the first.
We can easily do the same steps in reverse order: first you square it, and I then look for the cosine of the resulting number: . It’s easy to guess that the result will almost always be different. An important feature of complex functions: when the order of actions changes, the function changes.
In other words, a complex function is a function whose argument is another function: .
For the first example, .
Second example: (same thing). .
The action we do last will be called "external" function, and the action performed first - accordingly "internal" function(these are informal names, I use them only to explain the material in simple language).
Try to determine for yourself which function is external and which internal:
Answers: Separating inner and outer functions is very similar to changing variables: for example, in a function
We change variables and get a function.
Well, now we will extract our chocolate bar and look for the derivative. The procedure is always reversed: first we look for the derivative of the outer function, then we multiply the result by the derivative of the inner function. In relation to the original example, it looks like this:
Another example:
So, let's finally formulate the official rule:
Algorithm for finding the derivative of a complex function:
It seems simple, right?
Let's check with examples:
Derivative of a function- the ratio of the increment of the function to the increment of the argument for an infinitesimal increment of the argument:
Basic derivatives:
Rules of differentiation:
The constant is taken out of the derivative sign:
Derivative of the sum:
Derivative of the product:
Derivative of the quotient:
Derivative of a complex function:
Algorithm for finding the derivative of a complex function:
Well, the topic is over. If you are reading these lines, it means you are very cool.
Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!
Now the most important thing.
You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.
The problem is that this may not be enough...
For what?
For successfully passing the Unified State Exam, for entering college on a budget and, MOST IMPORTANTLY, for life.
I won’t convince you of anything, I’ll just say one thing...
People who have received a good education earn much more than those who have not received it. This is statistics.
But this is not the main thing.
The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...
But think for yourself...
What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?
GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.
You won't be asked for theory during the exam.
You will need solve problems on time.
And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.
It's like in sports - you need to repeat it many times to win for sure.
Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!
You can use our tasks (optional) and we, of course, recommend them.
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If you don't like our tasks, find others. Just don't stop at theory.
“Understood” and “I can solve” are completely different skills. You need both.
Find problems and solve them!
Problem B9 gives a graph of a function or derivative from which you need to determine one of the following quantities:
The functions and derivatives presented in this problem are always continuous, making the solution much easier. Despite the fact that the task belongs to the section of mathematical analysis, even the weakest students can do it, since no deep theoretical knowledge is required here.
To find the value of the derivative, extremum points and monotonicity intervals, there are simple and universal algorithms - all of them will be discussed below.
Read the conditions of problem B9 carefully to avoid making stupid mistakes: sometimes you come across quite lengthy texts, but there are few important conditions that affect the course of the solution.
If the problem is given a graph of a function f(x), tangent to this graph at some point x 0, and it is required to find the value of the derivative at this point, the following algorithm is applied:
Let us note once again: points A and B must be looked for precisely on the tangent, and not on the graph of the function f(x), as often happens. The tangent line will necessarily contain at least two such points - otherwise the problem will not be formulated correctly.
Consider points A (−3; 2) and B (−1; 6) and find the increments:
Δx = x 2 − x 1 = −1 − (−3) = 2; Δy = y 2 − y 1 = 6 − 2 = 4.
Let's find the value of the derivative: D = Δy/Δx = 4/2 = 2.
Task. The figure shows a graph of the function y = f(x) and a tangent to it at the point with the abscissa x 0. Find the value of the derivative of the function f(x) at the point x 0 .
Consider points A (0; 3) and B (3; 0), find the increments:
Δx = x 2 − x 1 = 3 − 0 = 3; Δy = y 2 − y 1 = 0 − 3 = −3.
Now we find the value of the derivative: D = Δy/Δx = −3/3 = −1.
Task. The figure shows a graph of the function y = f(x) and a tangent to it at the point with the abscissa x 0. Find the value of the derivative of the function f(x) at the point x 0 .
Consider points A (0; 2) and B (5; 2) and find the increments:
Δx = x 2 − x 1 = 5 − 0 = 5; Δy = y 2 − y 1 = 2 − 2 = 0.
It remains to find the value of the derivative: D = Δy/Δx = 0/5 = 0.
From the last example, we can formulate a rule: if the tangent is parallel to the OX axis, the derivative of the function at the point of tangency is zero. In this case, you don’t even need to count anything - just look at the graph.
Sometimes, instead of a graph of a function, Problem B9 gives a graph of the derivative and requires finding the maximum or minimum point of the function. In this situation, the two-point method is useless, but there is another, even simpler algorithm. First, let's define the terminology:
In order to find the maximum and minimum points from the derivative graph, just follow these steps:
This scheme only works for continuous functions - there are no others in Problem B9.
Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−5; 5]. Find the minimum point of the function f(x) on this segment.
Let's get rid of unnecessary information and leave only the boundaries [−5; 5] and zeros of the derivative x = −3 and x = 2.5. We also note the signs:
Obviously, at the point x = −3 the sign of the derivative changes from minus to plus. This is the minimum point.
Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−3; 7]. Find the maximum point of the function f(x) on this segment.
Let's redraw the graph, leaving only the boundaries [−3; 7] and zeros of the derivative x = −1.7 and x = 5. Let us note the signs of the derivative on the resulting graph. We have:
Obviously, at the point x = 5 the sign of the derivative changes from plus to minus - this is the maximum point.
Task. The figure shows a graph of the derivative of the function f(x), defined on the interval [−6; 4]. Find the number of maximum points of the function f(x) belonging to the segment [−4; 3].
From the conditions of the problem it follows that it is sufficient to consider only the part of the graph limited by the segment [−4; 3]. Therefore, we build a new graph on which we mark only the boundaries [−4; 3] and zeros of the derivative inside it. Namely, points x = −3.5 and x = 2. We get:
On this graph there is only one maximum point x = 2. It is at this point that the sign of the derivative changes from plus to minus.
A small note about points with non-integer coordinates. For example, in the last problem the point x = −3.5 was considered, but with the same success we can take x = −3.4. If the problem is compiled correctly, such changes should not affect the answer, since the points “without a fixed place of residence” do not directly participate in solving the problem. Of course, this trick won't work with integer points.
In such a problem, like the maximum and minimum points, it is proposed to use the derivative graph to find areas in which the function itself increases or decreases. First, let's define what increasing and decreasing are:
Let us formulate sufficient conditions for increasing and decreasing:
Let us accept these statements without evidence. Thus, we obtain a scheme for finding intervals of increasing and decreasing, which is in many ways similar to the algorithm for calculating extremum points:
Task. The figure shows a graph of the derivative of the function f(x) defined on the interval [−3; 7.5]. Find the intervals of decrease of the function f(x). In your answer, indicate the sum of the integers included in these intervals.
As usual, let's redraw the graph and mark the boundaries [−3; 7.5], as well as zeros of the derivative x = −1.5 and x = 5.3. Then we note the signs of the derivative. We have:
Since the derivative is negative on the interval (− 1.5), this is the interval of decreasing function. It remains to sum all the integers that are inside this interval:
−1 + 0 + 1 + 2 + 3 + 4 + 5 = 14.
Task. The figure shows a graph of the derivative of the function f(x), defined on the interval [−10; 4]. Find the intervals of increase of the function f(x). In your answer, indicate the length of the largest of them.
Let's get rid of unnecessary information. Let us leave only the boundaries [−10; 4] and zeros of the derivative, of which there were four this time: x = −8, x = −6, x = −3 and x = 2. Let’s mark the signs of the derivative and get the following picture:
We are interested in the intervals of increasing function, i.e. such where f’(x) ≥ 0. There are two such intervals on the graph: (−8; −6) and (−3; 2). Let's calculate their lengths:
l 1 = − 6 − (−8) = 2;
l 2 = 2 − (−3) = 5.
Since we need to find the length of the largest of the intervals, we write down the value l 2 = 5 as an answer.
