Question

函數(shù)f（x）=2x+log₃x-8的零點(diǎn)有________個(gè)．

Answer 1

1
分析：由題意，判斷此函數(shù)的零點(diǎn)個(gè)數(shù)可轉(zhuǎn)化為兩個(gè)函數(shù)y=-2x+8，與y=log₃x的交點(diǎn)個(gè)數(shù)結(jié)合兩個(gè)函數(shù)的圖象得出兩函數(shù)圖象的交點(diǎn)個(gè)數(shù)，即可得到原函數(shù)零點(diǎn)的個(gè)數(shù)．
解答：解：函數(shù)f（x）=2x+log₃x-8的零點(diǎn)與兩個(gè)函數(shù)y=-2x+8與y=log₃x的交點(diǎn)個(gè)數(shù)相同
由右圖知，函數(shù)y=-2x+8與y=log₃x的圖象僅有一個(gè)交點(diǎn)
故函數(shù)f（x）=2x+log₃x-8的零點(diǎn)有1個(gè)
故答案為 1
點(diǎn)評(píng)：本題考查函數(shù)的零點(diǎn)的定義及其個(gè)數(shù)的判斷，解題的關(guān)鍵是理解函數(shù)的零點(diǎn)定義，依據(jù)定義將求零點(diǎn)個(gè)數(shù)的問題轉(zhuǎn)化為兩個(gè)函數(shù)交點(diǎn)個(gè)數(shù)的問題，屬于基本題