Extreme value is a local concept defined by. The function value of a certain point is the maximum or minimum value compared with the function value of its nearby points, which does not mean that the maximum or minimum value of a function is not unique in the whole domain of the function, that is, there can be more than one maximum or minimum value of a function in a certain interval or domain, that is, the maximum value of a function is not necessarily greater than the minimum value. The endpoint of the interval can't be an extreme point, so the point where the function gets the maximum and minimum may be inside the interval, or the idea that f(x0) is the maximum or minimum can be judged by the derivative at the endpoint of the interval: if it is satisfied and the signs of the derivatives on both sides are different, it is an extreme point; if both sides satisfy "Zuo Zheng right negative", it is a maximum point, which is the maximum value; If both sides of "left negative and right positive" are satisfied, it is the minimum point of, which is the step of finding the extreme value of function f(x) with the minimum value: (1) determine the defined interval of the function, find the derivative f'(x) (2) find the root of equation f'(x)=0 (3) use the point where the derivative of the function is 0, and if the left is negative in turn. If the left and right signs are both positive or negative, then f(x) has infinite value at this root and the continuous function in the closed interval must have maximum and minimum values in the world; A continuous function in an open interval does not necessarily have a maximum and a minimum. The maximum value of the function is obtained by comparing the function values in the whole definition domain, and the extreme value of the function is obtained by comparing the function values near the extreme point. A function has at most a maximum value and a minimum value within its defined interval, and there may be more than one extreme value of the function, or there may be no step of finding the maximum value of the function by using derivatives: (1) Finding the extreme value includes; ⑵ Compare the extreme values of sum and find the maximum value of the function. Example 1 Find the extreme value of the column function: (1); (2) Solution: (1) order, the stagnation point of12+0+0+↗ maximum ↗ minimum ↗ is the maximum value of the function; It is the minimum value of the function. (2) Set the stagnation point as-1/kloc-0+0-↘ maximum ↗ minimum ↘, and the minimum value =-3; When, max =- 1 value. Example 2 Let the natural logarithm be the base, A be the constant and), and find the value of X when taking the minimum value. Solution: Let (1) be given by table X (-∞, -2)-2f '(x)+↗ maximum value ↘ minimum value 0-0+f (x). Example 3 Find the nearest point on a parabola. Solution: Let it be a point on a parabola, then. Take the extreme value at the same time, so that. Youde is the only stagnation point. It is the minimum point when or. At this point, that is, the closest point on the parabola is (2,2). Example 4 let the function f(x) =-ax, where a > 0, and find the range of a, so that the function f(x) is monotone in the interval [0, +∞]. Analysis: To make f, +∞) should be positive or negative. Solution: f ′ (x) =-A. When x > 0, because a > 0, if and only if a≥ 1, f ′ (x) =-A is always less than 0 on [0, +∞), then f(x) is monotonous. That is, f'(x)> 0 (or < 0) monotonically increases (or decreases). In example 5, it is known that the function f(x) = ax3+bx2-3x obtains the extreme value at x = 1. (1) Discuss f( 1) and f (-65438). (2) Make the tangent of curve y=f(x) pass through point A (0, 16) and find the tangent equation. Solution: (1)F'(x)= 3ax 2+2bx-3. According to the meaning of the question, f' (1) = f.f' (x) = 3x2-3 = 3 (x+1) (x-/) Let f ′ (x) = 0, and x =- 1, x= 1. If x∑(-)+∞) is also a increasing function. If x ∈ (- 1, 1), then f'(x) is less than 0, so f(x) is a decreasing function on (-1, 1). So f (- 1). 16) is not on the curve. Let the tangent point be M(x0, y0), then the coordinates of point m satisfy y0 = x03-3x0. Because f' (x0) = 3 (x02- 1), the tangent equation is y-y0 = 3 (x02-65438). There is16-(x03-3x0) = 3 (x0 =-2-1) (0-x0), which is simplified as x03 =-8 and solved as x0 =-2. So the tangent point is m (-2, -2) and the tangent equation is 9x-y+650.