Instruction
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The notion of critical point functions are closely related to the notion of its derivative at that point. Namely, point is called critical if the derivative of a function it does not exist or is zero. Critical points are the internal points of the domain of definition of the function.
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To determine the critical points of this function, you must perform several actions: find the domain of the function, calculate derivative, find the domain of definition of the derivative function, find the point treatment of the derivative to zero, to prove the belonging of the points defining the original function.
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Example 1Определите critical points of the function y = (x - 3)2·(x-2).
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Reinitiate the domain of the function, in this case there are no restrictions: x ∈ (-∞; +∞);Compute the derivative y’. According to the rules of differentiation of product of two functions is: y’ = ((x - 3)2)’·(x - 2) + (x - 3)2·(x - 2)’ = 2·(x - 3)·(x - 2) + (x - 3)2·1. After opening the brackets it turns out the quadratic equation: y’ = 3·x2 – 16·x + 21.
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Find the definition of the derivative of the function: x ∈ (-∞; +∞).Solve the equation 3·x2 – 16·x + 21 = 0 to find which x the derivative becomes zero: 3·x2 – 16·x + 21 = 0.
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D = 256 – 252 = 4x1 = (16 + 2)/6 = 3; x2 = (16 - 2)/6 = 7/3.So, the derivative becomes zero at values of x equal to 3 and 7/3.
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Determine whether found point determining the original function. Since x (-∞; +∞), then both these points are critical.
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Example 2Определите critical points of the function y = x2 – 2/x.
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Rectiability function definition: x ∈ (-∞; 0) ∪ (0; +∞), since x is in the denominator.Calculate the derivative y’ = 2·x + 2/x2.
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The scope of the definition of the derivative function is the same as the original: x ∈ (-∞; 0) ∪ (0; +∞).Solve the equation 2·x + 2/x2 = 0:2·x = -2/x2 → x = -1.
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So the derivative vanishes at x = -1. Done necessary but not sufficient condition for criticality. Since x=-1 falls within the interval (-∞; 0) ∪ (0; +∞), this point is critical.
