1作差(商)法构造函数
试题中出简单基初等函数例f(x)x3g(x)ln x证明某取值范围等式f(x)≥g(x)成立时构造函数h(x)f(x)g(x)φ(x)g(x)f(x)证明h(x)min≥0φ(x)max≤0求值程中利导数外够说明g(x)>0(f(x)>0)前提构造函数h(x)f(x)g(x)φ(x)g(x)f(x)证明h(x)min≥1(0<φ(x)max≤1)
典例1 已知函数f(x)exax(a常数)图象y轴交点A曲线yf(x)点A处切线斜率1
(1)求a值函数f(x)极值
(2)证明x>0时x2
f(x)ex2x f '(x)ex2令f '(x)0xln 2
x
xln 2时 f(x)极值极值f(ln 2)eln 22ln 22ln 4 f(x)极值
(2)证明令g(x)exx2g'(x)ex2x(1)g'(x)≥f(ln 2)>0
g(x)增函数x>0时g(x)>g(0)1>0x2
函数f(x)ln x+12x2+ax(a∈R)g(x)ex+32x2意x∈(0+∞)总f(x)≤g(x)成立求实数a取值范围
解析 f(x)≤g(x)⇒exln x+x2≥axx>0a≤ex+x2lnxx意x>0恒成立
设φ(x)ex+x2lnxx(x>0)φ'(x)ex+2x1xx(ex+x2lnx)x2ex(x1)+lnx+(x+1)(x1)x2
∵x>0∴x∈(01)时φ'(x)<0φ(x)单调递减x∈(1+∞)时φ'(x)>0φ(x)单调递增
∴φ(x)≥φ(1)e+1∴a≤e+1
2拆分法构造函数
证明等式基初等函数通相相加形式组成时果直接求导导函数种扑朔迷离知措感觉时原等式合理拆分f(x)≤g(x)形式进证明f(x)max≤g(x)min时注意配合导数工具拆分程中定注意合理性握般利导数进行值分析拆分标准
典例2 设函数f(x)aexln x+bex1x曲线yf(x)点(1 f(1))处切线ye(x1)+2
(1)求ab
(2)证明 f(x)>1
解析 (1)函数f(x)定义域(0+∞)f '(x)aexlnx+1x+bex1(x1)x2
题意f(1)2f '(1)e解a1b2
(2)证明(1)知f(x)exln x+2ex1xf(x)>1等价xln x>xex2e
构造函数g(x)xln x(x>0)g'(x)1+ln xx∈01e时g'(x)<0x∈1e+∞时g'(x)>0g(x)01e单调递减1e+∞单调递增g(x)(0+∞)值g1e1e构造函数h(x)xex2e(x>0)h'(x)ex(1x)x∈(01)时h'(x)>0x∈(1+∞)时h'(x)<0h(x)(0+∞)值h(1)1e
综x>0时g(x)>h(x)f(x)>1
点拨 第(2)问证明直接构造函数h(x)exln x+2ex1x(x>0)求导易分析先等式exln x+2ex1x>1合理拆分xln x>xex2e分左右两边构造函数进达证明原等式目
点训练1:(2017山东2013分)已知函数f(x)13x312ax2a∈R
(1)a2时求曲线yf(x)点(3 f(3))处切线方程
(2)设函数g(x)f(x)+(xa)cos xsin x讨g(x)单调性判断极值极值时求出极值
解析 (1)题意知f '(x)x2axa2时 f(3)0 f '(x)x22x
f '(3)3曲线yf(x)点(3 f(3))处切线方程y3(x3)3xy90
(2)g(x)f(x)+(xa)cos xsin xg'(x)f '(x)+cos x(xa)sin xcos xx(xa)(xa)sin x
(xa)(xsin x)
令h(x)xsin xh'(x)1cos x≥0
h(x)R单调递增h(0)0x>0时h(x)>0
x<0时h(x)<0
①a<0时g'(x)(xa)(xsin x)
x∈(∞a)时xa<0g'(x)>0g(x)单调递增
x∈(a0)时xa>0g'(x)<0g(x)单调递减
x∈(0+∞)时xa>0g'(x)>0g(x)单调递增
xa时g(x)取极值极值g(a)16a3sin a
x0时g(x)取极值极值g(0)a
②a0时g'(x)x(xsin x)
x∈(∞+∞)时g'(x)≥0g(x)单调递增
g(x)(∞+∞)单调递增g(x)极值极值
③a>0时g'(x)(xa)(xsin x)
x∈(∞0)时xa<0g'(x)>0g(x)单调递增
x∈(0a)时xa<0g'(x)<0g(x)单调递减
x∈(a+∞)时xa>0g'(x)>0g(x)单调递增
x0时g(x)取极值极值g(0)a
xa时g(x)取极值极值g(a)16a3sin a
综述
a<0时函数g(x)(∞a)(0+∞)单调递增(a0)单调递减函数极值极值极值g(a)16a3sin a极值g(0)a
a0时函数g(x)(∞+∞)单调递增极值
a>0时函数g(x)(∞0)(a+∞)单调递增(0a)单调递减函数极值极值极值g(0)a极值g(a)16a3sin a
3换元法构造函数
典例3 已知函数f(x)ax2+xln x(a∈R)图象点(1 f(1))处切线直线x+3y0垂直
(1)求实数a值
(2)求证n>m>0时ln nln m>mnnm
解析 (1)f(x)ax2+xln xf '(x)2ax+ln x+1切线直线x+3y0垂直切线斜率3f '(1)32a+13a1
(2)证明证ln nln m>mnnm证lnnm>mnnm
需证lnnmmn+nm>0
令nmx已知n>m>0nm>1x>1构造函数g(x)ln x1x+x(x>1)g'(x)1x+1x2+1x∈(1+∞)g'(x)1x+1x2+1>0g(x)(1+∞)单调递增
gnm>g(1)0证lnnmmn+nm>0成立命题证
点拨 证等式等价变形lnnmmn+nm>0观察知nm进行换元进构造函数g(x)ln x1x+x(x>1)证明等式简化证明程中运算
点训练2:已知函数f(x)x2ln x
(1)求函数f(x)单调区间
(2)证明意t>0存唯stf(s)
(3)设(2)中确定s关t函数sg(t)证明t>e2时25
x变化时 f '(x) f(x)变化情况表
x
01e
1e
1e+∞
f '(x)
0
+
f(x)
↘
极值
↗
函数f(x)单调递减区间01e单调递增区间1e+∞
(2)证明0
(3)证明sg(t)(2)知tf(s)s>1lng(t)lntlnslnf(s)lnsln(s2lns)lns2lns+ln(lns)u2u+lnu
中uln s25
10u>2时F '(u)<0u>1F(u)≤F(2)<0
ln u
典例4 (2017课标全国Ⅱ2112分)设函数f(x)(1x2)ex
(1)讨f(x)单调性
(2)x≥0时 f(x)≤ax+1求a取值范围
解析 (1)f '(x)(12xx2)ex令f '(x)0x12x1+2
x∈(∞12)时 f '(x)<0
x∈(121+2)时 f '(x)>0
x∈(1+2+∞)时 f '(x)<0
f(x)(∞12)(1+2+∞)单调递减(121+2)单调递增
(2)f(x)(1+x)(1x)ex
a≥1时设函数h(x)(1x)exh'(x)xex<0(x>0)
h(x)[0+∞)单调递减h(0)1
h(x)≤1f(x)(x+1)h(x)≤x+1≤ax+1
00(x>0)
g(x)[0+∞)单调递增g(0)0ex≥x+1
0
取x054a12x0∈(01)(1x0)(1+x0)2ax010f(x0)>ax0+1
a≤0时取x0512
x0∈(01) f(x0)>(1x0)(1+x0)21≥ax0+1
综a取值范围[1+∞)
点训练3:
已知函数f(x)exxln xg(x)extx2+xt∈R中e然数底数
(1)求函数f(x)图象点(1f(1))处切线方程
(2)g(x)≥f(x)意x∈(0+∞)恒成立求t取值范围
解析 (1)f(x)exxln x知f'(x)eln x1f'(1)e1f(1)e
求切线方程ye(e1)(x1)y(e1)x+1
(2)∵f(x)exxln xg(x)extx2+xt∈R
∴g(x)≥f(x)意x∈(0+∞)恒成立等价extx2+xex+xln x≥0意x∈(0+∞)恒成立t≤ex+xex+xlnxx2意x∈(0+∞)恒成立
令F(x)ex+xex+xlnxx2F'(x)xex+ex2exxlnxx31x2ex+e2exxlnx
令G(x)ex+e2exxln xx∈(0+∞)
G'(x)ex2(xexex)x21xex(x1)2+exxx2>0意x∈(0+∞)恒成立
∴G(x)ex+e2exxln x(0+∞)单调递增G(1)0
∴x∈(01)时G(x)<0x∈(1+∞)时G(x)>0
x∈(01)时F'(x)<0x∈(1+∞)时F'(x)>0
∴F(x)(01)单调递减(1+∞)单调递增
∴F(x)≥F(1)1∴t≤1
t取值范围(∞1]
5转化法构造函数
典例5 设函数f(x)ln x+mxm∈R
(1)me(e然数底数)时求f(x)值
(2)讨函数g(x)f '(x)x3零点数
(3)意b>a>0f(b)f(a)ba<1恒成立求m取值范围
解析 (1)me时 f(x)ln x+ex(x>0)f '(x)xex2x∈(0e)时 f '(x)<0 f(x)(0e)单调递减x∈(e+∞)时 f '(x)>0 f(x)(e+∞)单调递增xe时 f(x)取极值值 f(e)ln e+ee2f(x)值2
(2)g(x)f '(x)x31xmx2x3(x>0)令g(x)0m13x3+x(x>0)
设φ(x)13x3+x(x≥0)φ'(x)(x1)(x+1)x∈(01)时φ'(x)>0φ(x)(01)单调递增x∈(1+∞)时φ'(x)<0φ(x)(1+∞)单调递减x1φ(x)唯极值点极值点x1φ(x)值点φ(x)值φ(1)23φ(0)0结合yφ(x)图象知
①m>23时函数g(x)零点②m23时函数g(x)零点③0
h'(x)1xmx21≤0(0+∞)恒成立m≥x2+xx122+14(x>0)恒成立m≥14仅x12时等号成立m取值范围14+∞
点拨:例第(3)问中利等式性质f(b)f(a)ba<1等价转化f(b)b
点训练4: 已知函数f(x)12x2+(1a)xaln x
(1)讨f(x)单调性
(2)设a<0∀x1x2∈(0+∞)|f(x1)f(x2)|≥4|x1x2|求a取值范围
解析 (1)f(x)定义域(0+∞)
f '(x)x+1aaxx2+(1a)xax(x+1)(xa)x
a≤0f '(x)>0时f(x)(0+∞)单调递增
a>0f '(x)0xa0
时f(x)(0a)单调递减(a+∞)单调递增
(2)妨设x1≤x2a<0(1)知 f(x)(0+∞)单调递增∴f(x1)≤f(x2)∀x1x2∈(0+∞)|f(x1)f(x2)|≥4|x1x2|等价∀x1x2∈(0+∞)4x1f(x1)≥4x2f(x2)(*)
令g(x)4xf(x)g'(x)4f '(x)4x+1aaxaxx+3+a
(*)式等价g(x)(0+∞)单调递减
∴g'(x)axx+3+a≤0意x∈(0+∞)恒成立
∴a≤x23xx+1意x∈(0+∞)恒成立∴a≤x23xx+1min
x23xx+1x+1+4x+15≥2(x+1)·4x+151仅x+14x+1x1时等号成立
∴a≤1a取值范围(∞1]
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