已知函數(shù)f(x)=x3+ax2+bx+4在(-∞,0)上是增函數(shù),在(0,1)上是減函數(shù).
(Ⅰ)求b的值;
(Ⅱ)當(dāng)x≥0時(shí),曲線y=f(x)總在直線y=a2x-4上方,求a的取值范圍.
【答案】
分析:(Ⅰ)由題意得:f(x)在(-∞,0)上是增函數(shù),在(0,1)上是減函數(shù),所以當(dāng)x=0時(shí),f(x)有極大值,即f′(x)=0,即b=0.
(Ⅱ)因?yàn)閒(x)在(-∞,0)上是增函數(shù),在(0,1)上是減函數(shù),所以
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,即a
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.因?yàn)榍€y=f(x)在直線y=a
2x-4的上方,設(shè)g(x)=(x
3+ax
2+4)-(a
2x-4),
所以在x∈[0,+∝)時(shí),g(x)≥0恒成立.用導(dǎo)數(shù)求函數(shù)g(x)的最小值為g(-a),保證其大于0即可.
解答:解:(Ⅰ)∵f(x)=x
3+ax
2+bx+4,
∴f′(x)=3x
2+2ax+b.
∵f(x)在(-∞,0)上是增函數(shù),在(0,1)上是減函數(shù),
∴當(dāng)x=0時(shí),f(x)有極大值,即f′(x)=0,
∴b=0.
(Ⅱ)f′(x)=3x
2+2ax=x(3x+2a),
∵f(x)在(-∞,0)上是增函數(shù),在(0,1)上是減函數(shù),
∴
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,即a
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.
∵曲線y=f(x)在直線y=a
2x-4的上方,
設(shè)g(x)=(x
3+ax
2+4)-(a
2x-4),
∴在x∈[0,+∝)時(shí),g(x)≥0恒成立.
∵g′(x)=3x
2+2ax-a
2=(3x-a)(x+a),
令g′(x)=0,兩個(gè)根為-a,
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,且
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,
x | (0,-a) | -a | (-a,+∞) |
g′(x) | - | | + |
g(x) | 單調(diào)遞減 | 極小值 | 單調(diào)遞增 |
∴當(dāng)x=-a時(shí),g(x)有最小值g(-a).
令g(-a)=(-a
3+a
3+4)-(-a
3-4)>0,
∴a
3>-8,由
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,
∴-2<a
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.
點(diǎn)評(píng):解決此類問題的關(guān)鍵是將不等式在某個(gè)區(qū)間上恒成立問題轉(zhuǎn)化為函數(shù)在該區(qū)間上的最值問題,再利用導(dǎo)數(shù)求函數(shù)的最值,這也是高考考查的熱點(diǎn)之一.