Instruction

1

Find the derivative of the function. The derivative measures the rate of change of a function at a point is defined as the limit of the ratio of the increment function to increment of argument, which tends to zero. To identify it, use the table of derivatives. For example, the derivative of the function y = x3 is equal to y’ = x2.

2

Paranaita this derivative to zero (in this case, x2=0).

3

Find the value of the variable in the expression. It will be the values at which this derivative is equal to 0. To do this, substitute in the expression of arbitrary numbers instead of x in which the entire expression will become zero. For example:

2-2x2= 0

(1-x)(1+x) = 0

x1= 1, x2 = -1

2-2x2= 0

(1-x)(1+x) = 0

x1= 1, x2 = -1

4

The values obtained apply to direct coordinate and calculate the sign of the derivative for each of the periods. On the coordinate axis are marked points, which are taken as the starting point. To calculate the value of on intervals, substitute arbitrary values, appropriate criteria. For example, for the previous function to the interval -1 to choose a value of -2. On the interval from -1 to 1, you can choose 0, and for values greater than 1, select 2. Substitute these figures in the derivative and determine the sign of the derivative. In this case the derivative with x = -2 is equal to -0,24, i.e. negative on this interval will be a minus sign. If x=0, the value will be 2, and so on this interval is a positive sign. If x=1 then the derivative will also be equal to -0,24 and therefore is negative.

5

If the passage through the point on the coordinate axis the derivative changes its sign from minus to plus, it is a minimum point, and if from plus to minus, it is a maximum point.

Useful advice

For finding the derivative, there are online services that calculate the desired values, and output the result. On these sites you can find the derivative to 5 order.

# Advice 2 : How to find the least value of the function

The research function not only helps in plotting functions, but sometimes allows you to extract useful information on the functions, without resorting to the graphical image. So not necessarily to build a graph to find the smallest value of the function at some point.

Instruction

1

Imagine you are given the equation of the function y = f(x). The function is continuous and defined on the interval [a; b]. It is necessary to find the smallest value of the function on this interval. For example, consider the function f(x) = 3x2 + 4x3 + 1 on the interval [-2; 1]. Our f(x) is continuous and defined on the entire number line, and hence at a predetermined interval.

2

Find the first derivative of the function in variable x: f'(x). In our case, we get: f'(x) = 3*2x + 4*3x2 = 6x + 12x2.

3

Determine the points at which f'(x) equal to zero or not can be determined. In our example, f'(x) exists for all x, equate it to zero: 6x + 12x2 = 0 or 6x(1 + 2x) = 0. It is obvious that the product becomes zero if x = 0 or 1 + 2x = 0. Therefore, f'(x) = 0 when x = 0, x = -0,5.

4

Determine found among those points that belong to a given interval [a; b]. In our example, both points belong to the segment [-2; 1].

5

It remains to calculate the function values at the points of zero derivative, as well as at the ends of the segment. The smallest will be the smallest value of the function.

Compute function values at x = -2 and -0.5, 0, and 1.

f(-2) = 3*(-2)2 + 4*(-2)3 + 1 = 12 - 32 + 1 = -19

f(-0,5) = 3*(-0,5)2 + 4*(-0,5)3 + 1 = 3/4 - 1/2 + 1 = 1,25

f(0) = 3*02 + 4*03 + 1 = 1

f(1) = 3*12 + 4*13 + 1 = 3 + 4 + 1 = 8

Thus, the smallest value of the function f(x) = 3x2 + 4x3 + 1 on the interval [– 2; 1] is f(x) = -19, it is achieved at the left end of the segment.

Compute function values at x = -2 and -0.5, 0, and 1.

f(-2) = 3*(-2)2 + 4*(-2)3 + 1 = 12 - 32 + 1 = -19

f(-0,5) = 3*(-0,5)2 + 4*(-0,5)3 + 1 = 3/4 - 1/2 + 1 = 1,25

f(0) = 3*02 + 4*03 + 1 = 1

f(1) = 3*12 + 4*13 + 1 = 3 + 4 + 1 = 8

Thus, the smallest value of the function f(x) = 3x2 + 4x3 + 1 on the interval [– 2; 1] is f(x) = -19, it is achieved at the left end of the segment.

# Advice 3 : How to find the greatest least value of a function

Outstanding German mathematician Carl Weierstrass proved that for each continuous function on the interval, there are its maximum and minimum value on this interval. The problem of determining the largest and smallest values of the function has a wide application value in Economics, mathematics, physics and other Sciences.

You will need

- a blank sheet of paper;
- a pen or pencil;
- the textbook on higher mathematics.

Instruction

1

Let the function f(x) is continuous and defined on a given interval [a; b] and has on it some (finite) number of critical points. First, find the derivative function f'(x) h.

2

Equate the derivative of the function to zero to determine the critical points of the function. Do not forget to specify the point at which the derivative does not exist, they are also critical.

3

Of the many found the critical points we select those which belong to the segment [a; b]. The computed values of the function f(x) at these points and at the ends of the segment.

4

From the set of found values of the function to select the maximum and minimum values. This is the desired maximum and minimum values of the function.

# Advice 4 : How to find critical points functions

When plotting functions it is necessary to determine the points of maxima and minima, intervals of monotony of the function. To answer these questions the first thing you need to find the critical points, i.e. such points determine the function at which the derivative does not exist or is zero.

You will need

- The ability to find the derivative of the function.

Instruction

1

Find the domain of definition D(x) of the function y=ƒ(x), as all research functions are carried out in the interval where the function makes sense. If you examine the function on some interval (a; b), check that this interval belonged to the domain of definition D(x) of the function ƒ(x). Check the function ƒ(x) continuous in the interval (a; b). That is, lim(ƒ(x)) for x tending to each point x0 of the interval (a; b) must be equal to ƒ(x0). Also, the function ƒ(x) must be differentiable on this interval except possibly a finite number of points.

2

Calculate the first derivative ƒ'(x) the function ƒ(x). To do this, use the special table of derivatives of elementary functions and rules of differentiation.

3

Find the definition of the derivative ƒ'(x). Write down all the points that do not fall in the domain of the function ƒ'(x). Select from this set of points only those values that belong to the domain of definition D(x) of the function ƒ(x). This will be a critical point of the function ƒ(x).

4

Find all solutions of the equation ƒ'(x)=0. Select from these solutions, only those values that fall within the scope of the definition of D(x) of the function ƒ(x). These points will be critical points of a function ƒ(x).

5

Let's consider an example. Given a function ƒ(x)=2/3×x^3-2×x^2-1. The scope of this function is all number line. Find the first derivative ƒ'(x)=(2/3×x^3-2×x^2-1)’=(2/3×x^3)’−(2×x^2)’=2×x^2-4×x. The derivative ƒ'(x) is defined for any x value. Then solve the equation ƒ'(x)=0. In this case, 2×x^2-4×x=2×x×(x−2)=0. This equation is equivalent to the system of two equations: 2×x=0, i.e. x=0 and x−2=0, that is x=2. These two solutions belong to the field of determining the function ƒ(x). Thus, the function ƒ(x)=2/3×x^3-2×x^2-1, there are two critical points x=0 and x=2.

Note

The sign ^ denotes exponentiation, ' – taking the derivative.

# Advice 5 : How to find maximum value of function

Let us have some function, given analytically, i.e. expression of the form f(x). You need to investigate the function and calculate the maximum value that it takes on a given interval [a, b].

Instruction

1

First of all you need to establish whether the given function on the whole interval [a, b] and if it has break points, what kind of these breaks. For example, the function f(x) = 1/x does not have neither maximum nor minimum values on the interval [-1, 1] because the point x = 0 tends to plus infinity on the right and negative infinity on the left.

2

If the given function is linear, i.e. given by the equation of the form y = kx + b, where k ≠ 0, it is in all its scope is monotone increasing if k > 0; and decreases monotonically if k < 0. Therefore, its maximum value for any given segment will be f(b), if k > 0; and f(a), if k < 0.

3

The next step is the study of the function at the extrema. Even if it is determined that f(a) > f(b) (or Vice versa), the function may reach a large value at the point of max.

4

To find the maximum point, it is necessary to resort to the help of the derivative. We know that if at point x0 the function f(x) has a extremum (i.e. maximum, minimum or stationary point), then its derivative f'(x) at that point vanishes: f'(x0) = 0.

To determine which of three kinds of extremum is detected point, it is necessary to investigate the behavior of the derivative in its vicinity. If it changes sign from plus to minus, that is, monotonically decreasing, then the point of the original function has a maximum. If the derivative changes sign from minus to plus, i.e., monotonically increases, at the point of the original function has a minimum. If, finally, the derivative does not change sign, then x0 is a stationary point for the original function.

To determine which of three kinds of extremum is detected point, it is necessary to investigate the behavior of the derivative in its vicinity. If it changes sign from plus to minus, that is, monotonically decreasing, then the point of the original function has a maximum. If the derivative changes sign from minus to plus, i.e., monotonically increases, at the point of the original function has a minimum. If, finally, the derivative does not change sign, then x0 is a stationary point for the original function.

5

In cases where to compute the signs of the derivative in the vicinity of the point difficult, you can use the second derivative f"(x) and determine the sign of this function at the point x0:

- if f"(x0) > 0, the found minimum point;

- if f"(x0) < 0, the found maximum point;

finally, if f"(x0) = 0, then the stationary point.

- if f"(x0) > 0, the found minimum point;

- if f"(x0) < 0, the found maximum point;

finally, if f"(x0) = 0, then the stationary point.

6

For the final solution of the problem is necessary to select the maximum of the values of the function f(x) at the endpoints of the interval and all found high.

# Advice 6 : How to find the minimum value of the function

The need to find the minimum

**value**of a mathematical**function**is a practical interest in the solution of applied tasks, for example, in the economy. Great**value**for business is to minimize losses.Instruction

1

To find the minimum

**value****of the function**, you need to determine what value of the argument x0 to run the inequality y(x0) ≤ y(x), where x ≠ x0. Usually this problem is solved at a certain interval or in the whole region of values**of the function**, if one is not specified. One aspect of the solution is the location of the stationary points.2

A stationary point is called

**a value**argument, in which the derivative**of the function**vanishes. According to the Fermat's theorem if a differentiable function takes the extreme**value**at some point (in this case a local minimum), then this point is stationary.3

The minimum

**value of**a function often accepts it at this point, however, it is possible to determine not always. Moreover, it is not always possible to say with certainty what is the minimum**of the function**or it takes an infinitely small**value**. Then, as a rule, find the limit to which it tends when descending.4

In order to determine the minimum

**value****of the function**, you need to perform a sequence of actions, consisting of four stages: finding domain**of a function**, obtaining stationary points, analysis values**of the function**at these points and at the ends of the interval, identifying a minimum.5

So, imagine you are given some function y(x) on the interval with boundary points A and B. Find the area of its definition and find out whether the interval is its subset.

6

Calculate the derivative

**of the function**. Paranaita the resulting expression to zero and find the roots of the equation. Check whether these stationary points in the interval. If not, then at the next stage, they are ignored.7

Consider the interval subject to boundary types: open, closed, combined or infinite. It depends on how you look for the minimum

**value**. For example, the interval [A, b] is a closed interval. Substitute them into the function and calculate values. Do the same with a stationary point. Select the minimum result.8

With open and infinite intervals the situation is more complicated. There will have to find one-sided limits that do not always give an unambiguous result. For example, for the interval with one closed and one-punctured boundary of [A, b) should find the function at x = A and one-sided limit lim y as x → -0.

# Advice 7 : How to find the maximum point of the function

The point of maximum of the function along with the points of minimum are called points of extremum. At these points the function changes behavior. The extrema are defined on bounded intervals and a numerical are always local.

Instruction

1

The process of finding a local extremum is called the study of the function and is done by examining the first and second derivative of the function. Before the start of the study verify that the interval parameter value belongs to the valid values. For example, for the function F=1/x the value of the argument x=0 is not allowed. Or for a function Y=tg(x) argument cannot have a value of x=90°.

2

Verify that the function Y is differentiable for all specified interval. Find the first derivative Y'. It is obvious that before reaching the points of local maximum, the function increases, and the transition through the maximum, the function becomes decreasing. The first derivative in its physical sense a measure of the rate of change of a function. While the function increases, the speed of this process is a positive value. When passing through the local maximum of the function begins to decrease, and the rate of changing the function becomes negative. The transition rate of change of the function through zero occurs at the point of local maximum.

3

Therefore, on a plot of increase of the function, its first derivative is positive for all values of the argument in this interval. And Vice versa — on a plot of decreasing function value of the first derivative is less than zero. At the point of local maximum value of the first derivative is zero. Obviously, to find the local maximum of the function, it is necessary to find a point h in which the first derivative of this function equal to zero. If any argument value for the observable part x<h derivative must be positive, and for x>h - negative.

4

To find x solve the equation Y'=0. The value of Y(h) is a local maximum if the second derivative of the function at this point is less than zero. Find the second derivative Y", substitute in the obtained expression the value of the argument x= h and compare the result with zero.

5

For example, the function Y=-x2+x+1 over the interval from -1 to 1 has continuous derivative Y'=-2x+1. When x=1/2 the derivative is equal to zero, and the transition through this point the derivative changes sign from "+" to "-". The second derivative Y"=-2. Build on the points of the graph of the function Y=-x2+x+1 and check whether the point with abscissa x=1/2 a local maximum at a specified numeric interval axis.