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 a range of definition of function in which the derivative does not exist or is equal to zero.
It is required to you
- Ability to find a function derivative.
Instruction
1. Find a range of definition of D (x) the y= functions of ƒ (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 a range of definition of D (x) function of ƒ (x). Check function of ƒ (x) for continuity in this interval (a; b). That is lim (ƒ (x)) at x x0 aspiring to each point from an interval (a; b) ƒ (x0) has to be equal. Also function of ƒ (x) has to be differentiated on this interval except for perhaps final number of points.
2. Calculate the first derivative of ƒ' (x) functions of ƒ (x). For this purpose use the special table of derivative elementary functions and rules of differentiation.
3. Find ƒ derivative range of definition' (x). Write out all points which did not get to a range of definition of function of ƒ' (x). Select only those values which belong to a range of definition of D (x) function of ƒ from this set of points (x). It will also be critical points of function of ƒ (x).
4. Find all solutions of the equation of ƒ' (x) =0. Choose only those values which get to a range of definition of D (x) functions of ƒ from these decisions (x). These points will also be critical points of function of ƒ (x).
5. Review an example. Let function of ƒ (x) =2/3×x^3−2×x^2−1 be given. Range of definition of this function all numerical straight line. Find the first derivative of ƒ' (x)= (2/3×x^3−2×x^2−1) '= (2/3×x^3) ’− (2×x^2) ’=2×x^2−4×x. ƒ derivative' (x) is defined at any value x. Then solve ƒ 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 a range of definition of function of ƒ (x). Thus, function of ƒ (x) =2/3×x^3−2×x^2−1 has two critical points of x=0 and x=2.