Question

17．函數(shù)f（x）=x+sinxcosx在區(qū)間[-$\frac{π}{4}$，$\frac{π}{4}$]上的值域是[-$\frac{π+2}{4}$，$\frac{π+2}{4}$]．

Answer 1

分析 求導(dǎo)并判斷導(dǎo)數(shù)的正負(fù)，從而確定函數(shù)的單調(diào)性，從而確定函數(shù)的值域．

解答 解：∵f（x）=x+sinxcosx，
∴f′（x）=1+cosxcosx-sinxsinx=2cos²x＞0，
∴f（x）=x+sinxcosx在區(qū)間[-$\frac{π}{4}$，$\frac{π}{4}$]上單調(diào)遞增，
∴-$\frac{π}{4}$-$\frac{\sqrt{2}}{2}$•$\frac{\sqrt{2}}{2}$≤x+sinxcosx≤$\frac{π}{4}$+$\frac{\sqrt{2}}{2}$•$\frac{\sqrt{2}}{2}$，
∴-$\frac{π+2}{4}$≤f（x）≤$\frac{π+2}{4}$，
故答案為：[-$\frac{π+2}{4}$，$\frac{π+2}{4}$]．

點(diǎn)評(píng) 本題考查了導(dǎo)數(shù)的綜合應(yīng)用及函數(shù)的單調(diào)性與值域的求法，屬于中檔題．