Critical points are one of the most important aspects of a research of function by means of a derivative and have a wide scope. They are used in differential and variation calculations, play a large role in physics and mechanics.

## Instruction

1. The concept of a critical point of function is closely connected with a concept of its derivative of this point. Namely, the point is called critical if the function derivative in it does not exist or is equal to zero. Critical points are internal points of a range of definition function.

2. To define critical points of this function, it is necessary to perform several operations: to find a function range of definition, to calculate its derivative, to find a range of definition of derivative function, to find points of the address of a derivative to zero, to prove accessory of the found points of a range of definition of initial function.

3. An example 1opredelite critical points of function y = (x - 3) ²\· (x-2).

4. ReshenieNaydite a function range of definition, in this case there are no restrictions: x ∈ (-∞; + ∞); Calculate derivative y’. By rules of differentiation of performing two functions is available: y’ = ((x - 3)²)’ · (x - 2) + (x - 3) ²\· (x - 2)’ = 2 · (x - 3) · (x - 2) + (x - 3) ²\· 1. After removal of brackets the quadratic equation turns out: y’ = 3·x \-16 · x + 21.

5. Find a range of definition of derivative function: x ∈ (-∞; + ∞). Solve the equation 3·x \-16 · x + 21 = 0 to find at what x the derivative addresses in zero: 3 · x \-16 · x + 21 = 0.

6. D = 256 – 252 = 4x1 = (16 + 2)/6 = 3; x2 = (16 - 2)/6 = 7/3. So, the derivative addresses in zero at the values x equal 3 and 7/3.

7. Define whether the found points of a range of definition of initial function belong. As x (-∞; + ∞), both of these points are critical.

8. An example 2opredelite critical points of function y = x² – 2/x.

9. Resheniyeoblast function definitions: x ∈ (-∞; 0) ∪ (0; + ∞) as x costs in a denominator. Calculate derivative y’ = 2·x + 2/x².

10. Range of definition of derivative function the same that at initial: x ∈ (-∞; 0) ∪ (0; + ∞). Solve the equation 2·x + 2/x² = 0:2·x = - 2/x² → x =-1.

11. So, the derivative addresses in zero at x =-1. The necessary, but insufficient criticality condition is executed. As x=-1 gets to an interval (-∞; 0) ∪ (0; + ∞), this point are critical.