分析:根據(jù)題意,令t=x
2+2x-3,先求函數(shù)y=
的定義域,又由二次函數(shù)的性質(zhì),可得當(dāng)x≤-3時(shí),t=x
2+2x-3為減函數(shù),當(dāng)x≥1時(shí),t=x
2+2x-3為增函數(shù),進(jìn)而可得函數(shù)y=
的單調(diào)遞減區(qū)間為(-∞,-3],分析選項(xiàng)可得答案.
解答:解:令t=x
2+2x-3,
對(duì)于函數(shù)y=
,有x
2+2x-3≥0,解可得x≤-3或x≥1,即其定義域?yàn)閧x|x≤-3或x≥1}
又由二次函數(shù)的性質(zhì),可得當(dāng)x≤-3時(shí),t=x
2+2x-3為減函數(shù),當(dāng)x≥1時(shí),t=x
2+2x-3為增函數(shù),
即當(dāng)x≤-3時(shí),函數(shù)y=
的單調(diào)遞減,即函數(shù)y=
的單調(diào)遞減區(qū)間為(-∞,-3],
分析選項(xiàng),可得A在(-∞,-3]中,
故選A.
點(diǎn)評(píng):本題考查函數(shù)的單調(diào)性的判斷,應(yīng)當(dāng)明確單調(diào)區(qū)間在函數(shù)的定義域中,故解題時(shí)首先要求出函數(shù)的定義域.