增區(qū)間為[3,+∞),減區(qū)間為(-∞,-1]
分析:先根據(jù)函數(shù)的解析式求得函數(shù)的定義域,進(jìn)而根據(jù)函數(shù)y=x
2-2x-3的單調(diào)區(qū)間求得函數(shù)f(x)的單調(diào)區(qū)間.
解答:要使根號(hào)有意義需x
2-2x-3≥0
解得x≥3或x≤-1
故函數(shù)的定義域?yàn)閧x|x≥3或x≤-1}
對(duì)于函數(shù)y=x
2-2x-3=(x-1)
2-4
當(dāng)x≥1時(shí)函數(shù)單調(diào)性增,x≤1時(shí),函數(shù)單調(diào)減
∴f(x)=
當(dāng)x≥3時(shí)單調(diào)增,當(dāng)x≤-1時(shí)函數(shù)單調(diào)減
故函數(shù)的增區(qū)間為[3,+∞),減區(qū)間為(-∞,-1]
故答案為:增區(qū)間為[3,+∞),減區(qū)間為(-∞,-1]
點(diǎn)評(píng):本題主要考查了函數(shù)單調(diào)性及其區(qū)間.解題的時(shí)候一定要注意函數(shù)的定義域問(wèn)題.