《例谈抽象函数解法（Abstract the solution to abstract functions）.doc》由会员分享，可在线阅读，更多相关《例谈抽象函数解法（Abstract the solution to abstract functions）.doc（15页珍藏版）》请在三一文库上搜索。
1、例谈抽象函数解法（Abstract the solution to abstract functions）My well-organized documentAbstract: the solution to abstract functionsThe function is the hot spot of the annual college entrance examinationAnd the application of abstract function is one of the difficult pointsAn abstract function is a function
2、that does not give a specific function resolution or imageBut some properties or algorithms of function satisfaction are givenSuch function tests can comprehensively examine students understanding of function concepts and their ability of algebraic reasoning and reasoningIt can also examine students
3、 understanding and acceptance of mathematical symbol languageAnd understanding of general and special relationshipsSo the propositional favorIn the last few years, the high exam question is constantly appearingBecause abstract functions have some abstractionIts nature is hiddenSo students are more a
4、fraid of abstract functionsActually,A large number of abstract functions are abstracted from the basic functions learned in middle schoolWhen the problem solvingStart with the background of studying abstract functionsAccording to the nature of the abstract functionBy analogy, guess that it might be
5、some basic functionYou can always find a way to do thisThis article starts with this understandingIn this paper, some common types of abstract functions and their solutions are discussed1. Linear functional abstract functionsLinear functional abstract functionsIts a function that is abstracted by li
6、near functionsThe basic type: f (x + y) = f (x) + f (y) typical function: f (x) = kxSome data which are frequently used: f (0) = f (0 + 0) = f (0) + f (0) f (0) = 0F (0) = f (x - x) = f (x) + f (x) f (x) = f (x) -We know that f (x + y) = f (x) + f (y).F (4) = 16Lets take f of minus 2.F (4) = f (2 +
7、2) = f (2) + f (2) = 16 8 f (2) = f (2) = f (2) = 8We know that this is a function of f (x) on RF (2) = 1, f (x + y) = f (x) + f (y)The inequality: f (x) + f (x - 2) 3Analysis: it is assumed that f (x) is an abstract function of y = xAnd f of x is a monotone additionSolution: 3 = 3 f (2) = f (2) + f
8、 (2) + f (2) = f (6). f (x) + f (x 2) 3 = f (6) x + (x - 2) 6 x 0F of negative 1 is minus 2I want f of x in the intervalThe range of the topAnalysis: it is known from the problem setFunction f of x is an abstract functionSo we take the range of f of xThe key is to study its monotonicitySolution: set
9、 whennamely f (x) for increasing functionIn the condition ofSo y is equal to minus xtheAnd then lets say that x is equal to y is equal to 0F of 0 is equal to 2 f of 0. (0) = 0 fSo f of minus x is equal to f of x.F of x is a function f (1) = f (1) = 2F of minus 2 is equal to 2, f of minus 1 is minus
10、4 f (x) the range of values for - 422. Exponential function abstract functionExponential functional abstract functionsA function that is abstracted from an exponential functionBasic type: f (x + y) = f (x)? F of y is a typical function: f of x is equal to ax (a 0, a not 1).Some common data: f (x) do
11、es not equal 0, f (0) = 1 and f (x) 0Lets say that the domain of f of x is minus infinity+ up)Meet the conditions: existmakeFor any x and ySet upO:(1) f (0); (2) any value xThe positive and negative values of f (x)Analysis: it is assumed that f (x) is an abstract function of the exponential function
12、So I guess f of 0 is equal to 1 and f of x Solution: (1) to substitute y = 0theIf f of x is equal to 0For anyThere areIts a contradiction f (x) indicates a 0 f (0) = 1(2) let y = x not equal 0theAnd we know that f of x is not equal to 0 (2 x) 0 fThats f of x, So for any xF (x) constantThe function y
13、 = f (x) is defined on RFor any real number x, yF of x plus y is equal to f of x, right? F (y)And when the x 0 0 f (x) 1F (0) = 1, and when x 1To prove: f (x) decreases on RAnalysis: it is assumed that f (x) is an abstract function of the exponential functionSo I guess f of 0 is equal to 1And f (x)
14、When x 0 0 f (x) 1Guess 0 a 0 f (0) indicates 0 f (0) = 1F of 0 is f of x minus x is equal to f of x, right? F of minus x, f of minus x is equal to 1 over f of x.When x 0- x 0 0 f (x) 1 when x 1Proof: f (x) 0Set when x 0 0 f (x) 0 f (x2 - x1) = f (x2)? F (x1) = f (x2)/f (x1) f (x2) f (x) on the R3.
15、Logarithmic function abstract functionLogarithmic function abstract functionA function that is abstracted from a logarithm functionF (xy) = f (x) + f (y) typical function: f (x) = log a x (a 0, a not 1)F (1) = f (1) = f (1) = f (1) + f (1), f (1) = 0F of 1 is equal to f of x? F of x plus f of x is e
16、qual to 0, f of x is equal to minus f of x.Lets say that f of x is defined at (0Plus infinity, the monotonic functionmeetTo: (1) f (1);(2) if f (x) + f (x - 8) is less than or equal to 2Lets take the range of xAnalysis: lets guess that f (x) is an abstract function of the logarithm functionF (1) = 0
17、F of 9 is equal to 2Solution: (1) the (1) = 0 f(2)So f of x plus f of x minus 8 is less than or equal to f of 9.namely (x) is (0 fPlus infinityTherefore,Solution: 8 , y, , 0F of x is equal to f of x minus f of y.Inequality f (x - 6) -f () 1F (x - 6) -f () -f () 2f (4), f (x - 6) f (4) + f (4) + f ()
18、 f (x 6) f ()F of x is a function defined on R plus X-ray and 0, 6 x 6 0Solution to 6 0, x not 1).We have f of x for any real number, x, y, f of xy is equal to f of x times f of y.And f of minus 1 is 1F (27) = 9At that time(1) to determine the parity of f (x);(2) judge f (x) at 0The monotonicity of
19、+ infinityAnd give proof;(3) ifLets figure out the range of aAnalysis: from the set of questions, f (x) is an abstract function of a power functionSo you can guess that f of x is evenAnd in the 0Plus infinity is an augmented functionSolution: (1) y = -1F of minus x is equal to f of x times f of negative 1. f (1) = 1F of minus x is equal to f of x.F of x is even(2)When the f (x1) f (x2)So f of x is at 0Plus infinity is an augmented function(3) f (27) = 9againagainTherefore,one