Approximate value of the function increment
For sufficiently small values, the increment of the function is approximately equal to its differential, i.e. Dy » dy and therefore
Example 2. Find the approximate value of the increment of the function y= when the argument x changes from the value x 0 =3 to x 1 =3.01.
Solution. Let's use formula (2.3). To do this, let's calculate
X 1 - x 0 = 3.01 - 3 = 0.01, then
Du" 
.
Approximate value of a function at a point
In accordance with the definition of the increment of the function y = f(x) at the point x 0, when the argument Dx (Dx®0) is incremented, Dy = f(x 0 + Dx) - f(x 0) and formula (3.3) can be written
f(x 0 + Dx) » f(x 0) + . (3.4)
Special cases of formula (3.4) are the expressions:
(1 + Dx) n » 1 + nDx (3.4a)
ln(1 + Dx) » Dx (3.4b)
sinDx » Dx (3.4v)
tgDx » Dx (3.4g)
Here, as before, it is assumed that Dx®0.
Example 3. Find the approximate value of the function f(x) = (3x -5) 5 at point x 1 =2.02.
Solution. For calculations we use formula (3.4). Let's represent x 1 as x 1 = x 0 + Dx. Then x 0 = 2, Dx = 0.02.
f(2.02)=f(2 + 0.02) » f(2) +
f(2) = (3 × 2 - 5) 5 = 1
15 × (3 × 2 - 5) 4 = 15
f(2.02) = (3 × 2.02 - 5) 5 » 1 + 15 × 0.02 = 1.3
Example 4. Calculate (1.01) 5 , , ln(1.02), ln .
Solution
1. Let's use formula (3.4a). To do this, let's imagine (1.01) 5 in the form (1+0.01) 5.
Then, assuming Dx = 0.01, n = 5, we get
(1.01) 5 = (1 + 0.01) 5 » 1 + 5 × 0.01 = 1.05.
2. Presenting 1/6 in the form (1 - 0.006), according to (3.4a), we obtain
(1 - 0.006) 1/6 » 1 + 
.
3. Taking into account that ln(1.02) = ln(1 + 0.02) and assuming Dx=0.02, using formula (3.4b) we obtain
ln(1.02) = ln(1 + 0.02) » 0.02.
4. Likewise
ln = ln(1 - 0.05) 1/5 = 
.
Find approximate values of function increments
155. y = 2x 3 + 5 when the argument x changes from x 0 = 2 to x 1 = 2.001
156. y = 3x 2 + 5x + 1 with x 0 = 3 and Dx = 0.001
157. y = x 3 + x - 1 with x 0 = 2 and Dx = 0.01
158. y = ln x at x 0 = 10 and Dx = 0.01
159. y = x 2 - 2x at x 0 = 3 and Dx = 0.01
Find approximate values of functions
160. y = 2x 2 - x + 1 at point x 1 = 2.01
161. y = x 2 + 3x + 1 at x 1 = 3.02
162.y= 
at point x 1 = 1.1
163. y= at point x 1 = 3.032
164. y = at point x 1 = 3.97
165. y = sin 2x at point x 1 = 0.015
Calculate approximately
166. (1,025) 10 167. (9,06) 2 168.(1,012) 3
169. (9,95) 3 170. (1,005) 10 171. (0,975) 4
172. 173. 174.
175. 176. 177.
178.ln(1.003×e) 179.ln(1.05) 5 180.ln
181.ln0.98 182.ln 183.ln(e 2 ×0.97)
Function research and graphing
Signs of monotonicity of a function
Theorem 1 (necessary condition increasing (decreasing) function) . If the differentiable function y = f(x), xО(a; b) increases (decreases) on the interval (a; b), then for any x 0 О(a; b).
Theorem 2 (sufficient condition increasing (decreasing) function) . If the function y = f(x), xО(a; b) has a positive (negative) derivative at each point of the interval (a; b), then this function increases (decreases) on this interval.
Extrema of the function
Definition 1. A point x 0 is called a maximum (minimum) point of the function y = f(x) if for all x from some d-neighborhood of the point x 0 the inequality f(x) is satisfied< f(x 0) (f(x) >f(x 0)) for x ¹ x 0 .
Theorem 3 (Fermat) (a necessary condition for the existence of an extremum) . If point x 0 is the extremum point of the function y = f(x) and at this point there is a derivative, then
Theorem 4 (the first sufficient condition for the existence of an extremum) . Let the function y = f(x) be differentiable in some d-neighborhood of the point x 0 . Then:
1) if the derivative, when passing through the point x 0, changes sign from (+) to (-), then x 0 is the maximum point;
2) if the derivative, when passing through the point x 0, changes sign from (-) to (+), then x 0 is the minimum point;
3) if the derivative does not change sign when passing through the point x 0, then at the point x 0 the function does not have an extremum.
Definition 2. The points at which the derivative of a function vanishes or does not exist are called critical points of the first kind.
using the first derivative
1. Find the domain of definition D(f) of the function y = f(x).
3. Find critical points first kind.
4. Place critical points in the domain of definition D(f) of the function y = f(x) and determine the sign of the derivative in the intervals into which the critical points divide the domain of definition of the function.
5. Select the maximum and minimum points of the function and calculate the function values at these points.
Example 1. Examine the function y = x 3 - 3x 2 for an extremum.
Solution. In accordance with the algorithm for finding the extremum of a function using the first derivative, we have:
1. D(f): xО(-¥; ¥).
2. 
.
3. 3x 2 - 6x = 0 Þ x = 0, x = 2 - critical points of the first kind.
Derivative when passing through the point x = 0
changes sign from (+) to (-), therefore it is a point
Maximum. When passing through the point x = 2, the sign changes from (-) to (+), therefore this is the minimum point.
5. y max = f(0) = 0 3 × 3 × 0 2 = 0.
Maximum coordinates (0; 0).
y min = f(2) = 2 3 - 3 × 2 2 = -4.
Minimum coordinates (2; -4).
Theorem 5 (second sufficient condition for the existence of an extremum) . If the function y = f(x) is defined and twice differentiable in some neighborhood of the point x 0, and , then at the point x 0 the function f(x) has a maximum if and a minimum if .
Algorithm for finding the extremum of a function
using the second derivative
1. Find the domain of definition D(f) of the function y = f(x).
2. Calculate the first derivative
On the one hand, calculating the differential is much simpler than calculating the increment; on the other hand, dy≈∆y and the error allowed in this case can be made arbitrarily small by reducing ∆x. These circumstances make it possible in many cases to replace ∆y with the value dy. From the approximate equality dy≈∆y, taking into account that ∆y = f(x) – f(x 0), and dy=f'(x 0)(x-x 0), we obtain f(x) ≈ f(x 0) + f'(x 0)(x – x 0), where x-x 0 = ∆x.
Example. Calculate.
Solution. Taking the function, we have: 
. Assuming x 0 =16 (we choose ourselves so that the root is extracted), ∆x = 0.02, we get .
Example. Calculate the value of the function f(x) = e x at point x=0.1.
Solution. For x 0 we take the number 0, that is, x 0 =0, then ∆x=x-x 0 =0.1 and e 0.1 ≈e 0 + e 0 0.1 = 1+0.1 = 1.1. According to the table, e 0.1 ≈1.1052. The error was minor.
Let's note one more thing important property differential. The formula for finding the differential dy=f’(x)dx is correct as in the case when x is an independent variable, and in the case when x– function of a new variable t. This property of a differential is called the invariance property of its form. For example, for the function y=tg(x) the differential will be written in the form 
regardless of whether x
independent variable or function. In case x– the function is specifically specified, for example x=t 2 , then the calculation of dy can be continued, for which we find dx=2tdt and substitute it into the previously obtained expression for dy: 
.
If instead of formula (2) we used the non-invariant formula (1), then in the case where x is a function, we could not continue the calculation of dy in a similar way, since ∆x, generally speaking, does not coincide with dx.
Consider the widespread problem on approximate calculation of the value of a function using a differential.
Here and further we will talk about first-order differentials; for brevity, we will often simply say “differential”. The problem of approximate calculations using differentials has a strict solution algorithm, and, therefore, no special difficulties should arise. The only thing is that there are small pitfalls that will also be cleaned up. So feel free to dive in head first.
In addition, the section contains formulas for finding the absolute and relative errors of calculations. The material is very useful, since errors have to be calculated in other problems.
To successfully master the examples, you need to be able to find derivatives of functions at least at an intermediate level, so if you are completely at a loss with differentiation, please start with finding the derivative at a point and with finding the differential at the point. From technical means you will need a micro calculator with various mathematical functions. You can use the capabilities of MS Excel, but in this case it is less convenient.
The lesson consists of two parts:
– Approximate calculations using the differential value of a function of one variable at a point.
– Approximate calculations using full differential values of a function of two variables at a point.
The task under consideration is closely related to the concept of differential, but since we do not yet have a lesson on the meaning of derivatives and differentials, we will limit ourselves to a formal consideration of examples, which is quite enough to learn how to solve them.
Approximate calculations using the differential of a function of one variable
In the first paragraph, the function of one variable rules. As everyone knows, it is denoted by y or through f(x). For this task it is much more convenient to use the second notation. Let's go straight to popular example, which often occurs in practice:
Example 1
Solution: Please copy the working formula for approximate calculation using differential into your notebook:
Let's start to figure it out, everything is simple here!
The first step is to create a function. According to the condition, it is proposed to calculate cube root from among: , therefore corresponding function has the form: .
We need to use the formula to find the approximate value.
Let's look at left side formulas, and the thought comes to mind that the number 67 must be represented in the form. What's the easiest way to do this? I recommend next algorithm: let's calculate given value on the calculator:
– it turned out to be 4 with a tail, this is an important guideline for the solution.
As x 0 select a “good” value, so that the root is removed completely. Naturally this meaning x 0 should be as close as possible to 67.
In this case x 0 = 64. Indeed, .
Note: When with selectionx 0 there is still a problem, just look at the calculated value (in this case 
), take the nearest integer part (in this case 4) and raise it to the required power (in this case ). As a result, the desired selection will be made x 0 = 64.
If x 0 = 64, then the increment of the argument: .
So, the number 67 is represented as a sum 
First we calculate the value of the function at the point x 0 = 64. Actually, this has already been done earlier:
The differential at a point is found by the formula:
– You can also copy this formula into your notebook.
From the formula it follows that you need to take the first derivative:
And find its value at the point x 0:
.
Thus:
Everything is ready! According to the formula:
The approximate value found is quite close to the value 4.06154810045 calculated using a microcalculator.
Answer:
Example 2
Calculate approximately, replacing the increments of the function with its differential.
This is an example for independent decision. Approximate sample finishing and answer at the end of the lesson. For beginners, I recommend first calculating exact value on a microcalculator to find out what number to take as x 0, and which one – for Δ x. It should be noted that Δ x V in this example will be negative.
Some may have wondered why this task is needed if everything can be calmly and more accurately calculated on a calculator? I agree, the task is stupid and naive. But I’ll try to justify it a little. Firstly, the task illustrates the meaning of the differential function. Secondly, in ancient times, a calculator was something like a personal helicopter in modern times. I myself saw how a computer the size of a room was thrown out of one of the institutes somewhere in 1985-86 (radio amateurs came running from all over the city with screwdrivers, and after a couple of hours only the case remained from the unit). We also had antiques in our physics department, although they were smaller in size – about the size of a desk. This is how our ancestors struggled with methods of approximate calculations. A horse-drawn carriage is also transport.
One way or another, the task remained in the standard course higher mathematics, and it will have to be solved. This is the main answer to your question =).
Example 3
Calculate approximately the value of a function using a differential 
at the point x= 1.97. Calculate a more accurate function value at a point x= 1.97 using a microcalculator, estimate the absolute and relative error calculations.
In fact, this task can easily be reformulated as follows: “Calculate the approximate value 
using a differential"
Solution: We use the familiar formula:
In this case, a ready-made function is already given: 
. Once again, I would like to draw your attention to the fact that to denote a function, instead of “game” it is more convenient to use f(x).
Meaning x= 1.97 must be represented in the form x 0 = Δ x. Well, it’s easier here, we see that the number 1.97 is very close to “two”, so it suggests itself x 0 = 2. And, therefore: .
Let's calculate the value of the function at the point x 0 = 2:
Using formula , let's calculate the differential at the same point.
We find the first derivative:
And its meaning at the point x 0 = 2:
Thus, the differential at the point:
As a result, according to the formula:
The second part of the task is to find the absolute and relative error of the calculations.
