Instruction

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A function y = F(x) is increasing on a particular interval if for any points x1 < x2 on this interval the condition F(x1) < F(x2). I.e. the greater the argument value, the greater the value of the function. For a decreasing function true F(x1) > F(x2), where x1 > x2 for any point on the interval.

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There are sufficient signs of ascending & descending functions, which are derived from the result of calculating the derivative. If the derivative of the function is positive for any point of the interval, the function is increasing, if negative, decreases.

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To find the intervals of increase and decrease functions, you need to find the area of its definition, calculate the derivative, solve the inequality F’(x) > 0 and F’(x) < 0, and then included in the resulting interval boundary points at which the function is continuous and defined to exclude those in which its value cannot be determined.

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Let's consider an example.

Find the intervals of increasing and decreasing function for y = (3·x2 + 2·x - 4)/x2.

Find the intervals of increasing and decreasing function for y = (3·x2 + 2·x - 4)/x2.

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Solution.

1. Find the domain of the function. Clearly, the expression standing in the denominator should always be different from zero. Therefore, the point 0 is excluded from the scope: the function is defined at x ∈ (-∞; 0)∪(0; +∞).

1. Find the domain of the function. Clearly, the expression standing in the denominator should always be different from zero. Therefore, the point 0 is excluded from the scope: the function is defined at x ∈ (-∞; 0)∪(0; +∞).

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2. Compute the derivative of the function:

y’(x) = ((3·x2 + 2·x - 4)’ ·x2 – (3·x2 + 2·x - 4) · (x2)’)/x^4 = ((6·x + 2) ·x2 – (3·x2 + 2·x - 4) ·2·x)/x^4 = (6·x3 + 2·x2 – 6·x3 – 4·x2 + 8·x)/x^4 = (8·x – 2·x2)/x^4 = 2· (4 - x)/x3.

y’(x) = ((3·x2 + 2·x - 4)’ ·x2 – (3·x2 + 2·x - 4) · (x2)’)/x^4 = ((6·x + 2) ·x2 – (3·x2 + 2·x - 4) ·2·x)/x^4 = (6·x3 + 2·x2 – 6·x3 – 4·x2 + 8·x)/x^4 = (8·x – 2·x2)/x^4 = 2· (4 - x)/x3.

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3. Solve the inequality y’ > 0 and y’ <0:

(4 - x)/x3 > 0;

(4 - x)/x3 < 0.

(4 - x)/x3 > 0;

(4 - x)/x3 < 0.

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4. The left part of the inequality has one real root x = 4 and goes to infinity at x = 0. Therefore, the value x = 4 is included in a period of increasing functions in the interval descending, and point 0 is not included anywhere.

Thus, the required function is increasing on the interval x ∈ (-∞; 0) ∪ [2; +∞) and decreases with x (0; 2].

Thus, the required function is increasing on the interval x ∈ (-∞; 0) ∪ [2; +∞) and decreases with x (0; 2].

# Advice 2 : How to find the intervals of increase and decrease

A function y=f(x) is increasing on some interval if for arbitrary x2>x1 f(x2)>f(x1). If thus f(x2)

You will need

- paper;
- - handle.

Instruction

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It is known that for an increasing function y=f(x) its derivative f’(x)>0 and accordingly f’(x)

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Example: find the intervals of monotonicity of y=(x^3)/(4-x^2). Solution. Function defined on the entire number line except x=2 and x=-2. Furthermore it is odd. Indeed, f(-x)=((-x)^3)/(4-(-x)^2)= -(x^3)/(4-x^2)=f(-x). This means that f(x) is symmetrical about the origin. Therefore, the study of the behavior of a function can be made only for positive values of x, and then finish the negative branch symmetrically positive.y’=(3(x^2)(4-x^2)+2x(x^3))/((4-x^2)^2)=(x^2)(12-x^2)/((4-x^2)^2).y’ does not exist at x=2 and x=-2, but it does not exist and the function itself.

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*Now we need to find intervals of monotony of the function. This should solve the inequality: (x^2)(12-x^2)/((4-x^2)^2)>0 or (x^2)(x-2sqrt3)(x+2sqrt3)((x-2)^2)((x+2)^2))0. Use the method of intervals when solving inequalities. Then get (see Fig.1).*

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Next, consider the behavior of functions on intervals, monotonicity, adding here all the information of the region of negative values of a numeric axis (because of symmetry all the information there is the opposite, including in sign).f’(x)>0 when –∞

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Example 2. Find the intervals

**of increase**and*decrease*of the function y=x+lnx/x.Solution. The domain of the function is x>0.y’=1+(1-lnx)/(x^2)=(x^2+1-lnx)/(x^2). The sign of the derivative when x>0 is completely determined by the bracket (x^2+1-lnx). Since x^2+1>lnx, y’>0. Thus, the function increases throughout its domain of definition.6

*Example 3. Find intervals of monotony of the function y’=x^4-2x^2-5.Solution. y’=4x^3-4x=4x(x^2-1)=4x(x-1)(x+1). Applying the method of intervals (see Fig.2), you need to find the intervals of positive and negative values of the derivative. Using method of intervals, you will be able to quickly determine that the intervals x0, the function increases.*

# Advice 3 : How to determine the intervals of monotonicity

An interval

**of monotonicity**of a function can be called interval in which the function either only increases or only decreases. A number of specific actions that will help to find such ranges for functions that are frequently required in algebraic problems of this kind.Instruction

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The first step in solving the task of identifying the intervals in which the function monotonically increases or decreases, will be the calculation of the scope of this function. For that, find all values of the arguments (the values on the x-axis), for which we can find the value of the function. Mark the points where there are gaps. Find the derivative of the function. Determine the expression that represents the derivative, Paranaita it to zero. After that it is necessary to find the roots of the resulting equation. Don't forget about the range of permissible values.

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The points at which the function exists, or where its derivative is equal to zero, represents the boundaries of the intervals

**of monotonicity**. These ranges, as well as the point between them should be sequentially in the table. Find the sign of the derivative functions in the resulting gaps. To do this, substitute in the expression corresponding to the derivative, any argument from the interval. If the result is positive, the function in the given range increases, otherwise decreases. Results are entered into the table.3

In the row indicating the derivative of the function f'(x) is written corresponding to the argument values symbol: "+" — if the derivative is positive" -" negative or "0" is equal to zero. In the next line note the monotony of the original expression. The up arrow corresponds to increasing, arrow down – decreasing order. Mark the points of the extremum of the function. This is the point at which the derivative is equal to zero. The extremum can be either a maximum point or a minimum point. If the previous plot functions increased, and the current decreases, then it is the maximum point. In the case where from a given point the function is decreasing, now increases is a minimum point. Fill in the table values of the function at the points of extremum.

# Advice 4 : How to find the intervals of monotonicity and extremum

The study of the behavior of a function with a complex dependence on the argument is carried out using a derivative. The nature of the change of the derivative can find critical points and areas of growth or decrease of a function.

Instruction

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On different parts of the numerical plane function behaves differently. When crossing the y-axis of the function changes sign, passing through zero value. The monotonous rise may be replaced by a decrease when passing functions through the critical point — the extremes. To find the extrema of the function, the point of intersection with the coordinate axes, the plots repetitive behavior — all of these tasks are the analysis of the behavior of the derivative.

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Before beginning the study of the behaviour of the function Y = F(x) consider the scope of permissible argument values. Take into consideration only those values of the independent variable "x" when the existence of feature Y.

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Check whether the given function differentiable on the interval of the numerical axis. Find the first derivative of the given function Y' = F'(x). If F'(x)>0 for all values of the argument, the function Y = F(x) on this segment is increasing. The converse is true: if the interval F'(x)<0, the phase function decreases monotonically.

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For finding extrema solve the equation F'(x)=0. Determine the argument value x₀ at which the first derivative equal to zero. If the function F(x) exists at x=h and equal Y₀=F(x₀), the resulting point is an extremum.

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To determine the found extremum point is a maximum or minimum of a function, compute the second derivative F"(x) the original function. Find the value of the second derivative at the point x₀. If F"(x₀ )>0, then x₀ - point minimum. If F"(x₀ )<0, x₀ is the maximum point of the function.