一次函数应用题例析（Analysis of application questions of primary function）.docVIP

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一次函数应用题例析（Analysis of application questions of primary function）.doc

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一次函数应用题例析（Analysis of application questions of primary function）

一次函数应用题例析（Analysis of application questions of primary function） A large circle represents what I have learned, but there are so many blanks outside the circle that means ignorance to me. And the bigger the circle, the longer the circumference of the circle, and the greater the gap between it and the outside world. This shows that I do not understand the place is still very big. Analysis of application questions of primary function Primary function is one of the key parts in junior high school mathematics Design a primary function model to solve practical problems Throughout much of the proposition of all ages. The study is a few comments on senior high school entrance examination questions For your information Image type Case 1 (2003 Guangxi) was in the fight against SARS A medical research institute developed a drug to prevent SARS. The effect of the drug was tested 1 hours after the prescribed dose was taken by adults The blood contains the highest dose Reach 5 micrograms per milliliter Then gradually decay To 8 hours, the blood contains an amount of 1.5 micrograms per milliliter. The amount of Y (micrograms) in the blood per milliliter varies with time x (hours), as shown in the adult prescribed dose: (1) were calculated for x = 1 When x is greater than 1 the function relationship between Y and X type; (2) if the amount of blood contained in each milliliter is 2 micrograms or more than 2 micrograms The prevention of SARS is effective How many hours is this effective time? Analysis of the background material involved, professional strong But as long as you read The knowledge of functions we have learned is not difficult to answer. The main information of the subject is given by the function picture An image is a broken line made up of two lines Think of it as a combination of two primary function images (1) when x is less than or equal to 1 Setting y=k1x. will be (1) 5) substitute Get k1=5. * y=5x. When x 1 Set y=k2x+b. to (1) 5) (8 1.5) substitute have to L (2)