解:(1)由x≥0時(shí),f(x)=x
2-2x,
當(dāng)x<0時(shí),-x>0,
∴f(-x)=x
2+2x
又函數(shù)f(x)為偶函數(shù),
∴f(x)=x
2+2x-------------3’
故函數(shù)的解析式為
-------------4’
函數(shù)圖象如下圖所示:--------------7’
(2)由函數(shù)的圖象可知,
函數(shù)f(x)的單調(diào)遞增區(qū)間為[-1,0]、[1,+∞)
單調(diào)遞減區(qū)間為(-∞,-1]、[0,1],
函數(shù)f(x)的值域?yàn)閇-1,+∞)------12’
分析:(1)當(dāng)x<0時(shí),-x>0,由已知中當(dāng)x≥0時(shí),f(x)=x
2-2x,及函數(shù)f(x)是定義在R上的偶函數(shù),可求出當(dāng)x<0時(shí)函數(shù)的解析式,進(jìn)而得到答案,再由二次函數(shù)的圖象畫法可得到函數(shù)的草圖;
(2)根據(jù)圖象下降對應(yīng)函數(shù)的單調(diào)遞減區(qū)間,圖象上升對應(yīng)函數(shù)的單調(diào)遞增區(qū)間,分析出函數(shù)值的取值范圍后可得到答案
點(diǎn)評:本題考查的知識點(diǎn)是函數(shù)圖象,函數(shù)的單調(diào)區(qū)間,函數(shù)的值域,是函數(shù)圖象和性質(zhì)的綜合應(yīng)用,難度中檔.