How to find critical points of function

How to find critical points of function

At creation of a function graph it is necessary to define points of a maximum and a minimum, intervals of monotony of function. To answer these questions first of all it is necessary to find critical points, that is such points of range of definition of function in which the derivative does not exist or it is equal to zero.

It is required to you

Ability to find a function derivative.

The sponsor of placement P&G Articles on the subject "How to Find Critical Points of Function" How to put square roots How to find a square diagonal How to find parabola top coordinates

Instruction

1

Find range of definition of D (x) functions y =? (x) as all researches of function are conducted in that interval where function makes sense. If you investigate function on some interval (a; b), check that this interval belonged to range of definition of D (x) function? (x). Check function? (x) on a continuity in this interval (a; b). That is lim (? (x)) at x x0 aspiring to each point from an interval (a; b) it has to be equal? (x0). Also function? (x) it has to be differentiated on this interval except for perhaps final number of points.

2

Calculate the first derivative?' (x) functions? (x). For this purpose use the special table of derivative elementary functions and rules of differentiation.

3

Find derivative range of definition?' (x). Write out all points which did not get to function range of definition?' (x). Select only those values which ranges of definition of D (x) functions belong from this set of points? (x). It also there will be critical points of function? (x).

4

Find all solutions of the equation?' (x) =0. Choose only those values which get to range of definition of D (x) functions from these decisions? (x). These points will also be critical points of function? (x).

5

Review an example. Let function be given? (x) =2/3? x^3? 2? x^2? 1. Range of definition of this function all numerical straight line. Find the first derivative?' (x) = (2/3? x^3? 2? x^2? 1)’ = (2/3? x^3)’? (2? x^2)’ =2? x^2? 4? x. Derivative?' (x) it is defined at any value x. Then solve the equation?' (x) =0. In this case 2? x^2? 4? x=2? x? (x? 2) =0. The system from two equations is equivalent to this equation: 2? x=0, that is x=0, and x? 2=0, that is x=2. These two decisions belong to function range of definition? (x). Thus, at function? (x) =2/3? x^3? 2? x^2? 1 there are two critical points of x=0 and x=2.