Write the given function f(x). Define its first derivative f’(x). The obtained expression of the derivative Paranaita to zero.
Solve the resulting equation. The roots of the equation will be the critical points of the function.
Determine the critical points of maximum or minimum are obtained roots. To do this, find the second derivative f’(x) from the original function. Substitute it in turns to the critical point and calculate the expression. If the second derivative of the function at the critical point is greater than zero, it will be a minimum point. Otherwise the maximum point.
Calculate the value of the original function to the obtained points of minimum and maximum. To do this, substitute their values in the expression for the function and calculate. The resulting number will determine the extremum of the function. Moreover, if the critical point was a maximum, the extremum of the function will also be high. Also, the minimum critical point, the function will reach its minimum extremum.
Advice 2 : How to find the extremum of function of two variables
By definition, the point M0(x0, y0) is called a point of local maximum (minimum) of a function of two variables z=f(x,y) if in some neighborhood of the point U(x0, y0), for any point M(x, y) is executed f(x,y)f(x0, y0)). These points are called the extrema of the function. In the text of the partial derivatives are denoted in accordance with Fig. 1.
Necessary condition of extremum is equal to zero partial derivatives of a function in x and y. Point M0(x0, y0), which in turn zero, both partial derivatives is called a stationary point of a function z=f(x, y).
Remark. Partial derivatives of the function z=f(x, y) may not exist at the point of extremum, so the points of extremum are not only stationary points, but the points at which partial derivatives do not exist (they correspond to the edge surface of the graphics function).
Now you can go to sufficient conditions for the existence of the extremum. If a differentiable function has an extremum, then it can only be at a stationary point. Sufficient conditions are formulated as follows: suppose that in some neighborhood of the stationary point (x0, y0) the function f(x, y) has continuous partial derivatives of second order. For example: (cm. Fig.2)
Then: a) if Q>0, the point (x0, y0) the function has an extremum, with f’(x0, y0)0) - local minimum; b) if Q
For finding the extremum of functions of two variables can suggest the following scheme: first, are the stationary points of the function. Then these points are checked sufficient conditions. If the function in some points does not have partial derivatives in these points also can be an extremum, but sufficient conditions are no longer applicable.
Example. Find extrema of the function z=x^3+y^3-xy.Solution. Find the stationary points of the function (see Fig. 3):
The latter system gives the stationary point (0, 0) and (1/3, 1/3). It is now necessary to check that sufficient conditions. Find the second derivative and the stationary points Q(0,0 ) and Q(1/3, 1/3) (see figure 4):
Since Q(0, 0)0, consequently, at the point (1/3, 1/3) is the extremum. Given the fact that the second derivative (xx) to (1/3, 1/3) is greater than zero, you must make a decision that this point is a minimum.
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.
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.
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.
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.
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 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.
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.
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.
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.
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.
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.