Question

1．已知函數(shù)f（x）=x³-3$\sqrt{2}$x²+3x+1，討論函數(shù)的單調(diào)性．

Answer 1

分析 求函數(shù)的導(dǎo)數(shù)，利用函數(shù)單調(diào)性和導(dǎo)數(shù)之間的關(guān)系進(jìn)行判斷即可．

解答 解：函數(shù)的導(dǎo)數(shù)f′（x）=3x²-6$\sqrt{2}$x+3，
判別式△=（6$\sqrt{2}$）²-4×3×3=72-36=36，
由f′（x）=3x²-6$\sqrt{2}$x+3=0得方程的根為x₁=$\frac{6\sqrt{2}+\sqrt{36}}{6}=\frac{6\sqrt{2}+6}{6}$=1+$\sqrt{2}$，或x₂=$\frac{6\sqrt{2}-\sqrt{36}}{6}$=$\sqrt{2}$-1，
由f′（x）＞0得x＞1+$\sqrt{2}$或x＜$\sqrt{2}$-1，此時(shí)函數(shù)單調(diào)遞增，即函數(shù)單調(diào)遞增區(qū)間為（-∞，$\sqrt{2}$-1），（$\sqrt{2}$+1，+∞），
由f′（x）＜0得$\sqrt{2}$-1＜x＜$\sqrt{2}$+1，此時(shí)函數(shù)單調(diào)遞減，即函數(shù)單調(diào)遞減區(qū)間為（$\sqrt{2}$-1，$\sqrt{2}$+1）．

點(diǎn)評 本題主要考查函數(shù)單調(diào)性的判斷，根據(jù)函數(shù)單調(diào)性和導(dǎo)數(shù)之間的關(guān)系是解決本題的關(guān)鍵．