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 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.