已知函數(shù)f(x)=x2-2mx+2-m
(1)若不等式f(x)≥-mx+2在R上恒成立,求實(shí)數(shù)m的取值范圍
(2)設(shè)函數(shù)f(x)在[0,1]上的最小值為g(m),求g(m)的解析式及g(m)=1時(shí)實(shí)數(shù)m的值.
解:(1)由題意知,f(x)≥-mx在R上恒成立,
即x
2-mx+2-m≥0恒成立,
∴△=m
2+4m-8≤0,
解得-2-2
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.
∴實(shí)數(shù)m的取值范圍是[-2-2
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,-2+2
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].
(2)函數(shù)f(x)=x
2-2mx+2-m的對(duì)稱軸為x=m,
①當(dāng)m<0時(shí),
函數(shù)f(x)在[0,1]上的最小值g(m)=f(0)=2-m.
②當(dāng)0≤m≤1時(shí),
函數(shù)f(x)在[0,1]上的最小值g(m)=f(1)=-3m+3,
綜上所述,g(x)=
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,
∵g(m)=1,
∴m=
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.
分析:(1)由題意知,f(x)≥-mx在R上恒成立,即x
2-mx+2-m≥0恒成立,由此能求出實(shí)數(shù)m的取值范圍.
(2)函數(shù)f(x)=x
2-2mx+2-m的對(duì)稱軸為x=m,由此進(jìn)行分類討論,能夠求出g(m)的解析式及g(m)=1時(shí)實(shí)數(shù)m的值.
點(diǎn)評(píng):本題考查函數(shù)恒成立的應(yīng)用,解題時(shí)要認(rèn)真審題,仔細(xì)解答,注意分類討論思想和等價(jià)轉(zhuǎn)化思想的合理運(yùn)用.