Question

4．已知函數(shù)f（x）=$\left\{\begin{array}{l}{lo{g}_{\frac{1}{2}}x，x＞0}\\{-{x}^{2}-2x，x≤0}\\{\;}\end{array}\right.$，則不等式f（x）≤0的解集為{x|x≥1或x=0或x≤-2}．

Answer 1

分析 不等式f（x）≤0等價(jià)于$\left\{\begin{array}{l}{lo{g}_{\frac{1}{2}}x≤0}\\{x＞0}\end{array}\right.$或$\left\{\begin{array}{l}{-{x}^{2}-2x≤0}\\{x≤0}\end{array}\right.$，解得即可．

解答 解：函數(shù)f（x）=$\left\{\begin{array}{l}{lo{g}_{\frac{1}{2}}x，x＞0}\\{-{x}^{2}-2x，x≤0}\\{\;}\end{array}\right.$，則不等式f（x）≤0等價(jià)于$\left\{\begin{array}{l}{lo{g}_{\frac{1}{2}}x≤0}\\{x＞0}\end{array}\right.$或$\left\{\begin{array}{l}{-{x}^{2}-2x≤0}\\{x≤0}\end{array}\right.$，
解得x≥1或x=0或x≤-2，
故不等式的解集為{x|x≥1或x=0或x≤-2}
故答案為{x|x≥1或x=0或x≤-2}

點(diǎn)評(píng) 本題考查了分段函數(shù)和不等式的解法，培養(yǎng)了學(xué)生的運(yùn)算能力，屬于基礎(chǔ)題．