第四章微分中值定理与公式-总练习题

f(xf(x)0在(ab)证g(xexf(xg(ag(b0g在[ab](ab)可导根据Rollec(a,b),使得g(x)exf(xf(x0,即f(xf(x)决定常数A的范围,使方程3x48x36x224xA有四个不相等的实根解P(x3x48x36x224xP(x)12x324x212x12(x32x2x2)12[x2(x2)(x2)]12(x2)(x21)12(x2)(x1)(xx11,x21,x32.P(x1)19,P(1)13,P(2)3x48x36x224xA n设f(x)1x

n证当x0时f(x0,故f只有正根当n2k1证

f(x)

f(x存在abab,f(a0,f(b根据连续函数的中间值定理,存在x0(ab使得f(x0f(x)1xx2

x2k

x2k1x

0(x0),当x0时,f严格单调递减

2k x2k当n2k为偶数时,f(x)=1xx x10,x0x1f(x0x1,f(x0,f(1)是x0时的最小值,f(1)0,故当n为偶数时f(x)无实根.[a,b]上恒不等于零.证明u(x)在v(x)的相邻根之间必有一根,反之也对.即有u(x)与)的根互相交错地出现.试句举处满足上述条件的u(x)与[xx]上没有根则wu在[ab]连续w(xw(x0,由Rolle定理存在c[xx使得 w(c)uvuv(c)0,即(uvuv)(c)0,此与uvuv恒不等于零的假 .故在[x1,x2]上有证明:当x0时函数f(xarctanx单调递增且arctanx(tanhtanh tanh证f(xarctanx1

arctancosh2xsinhxcoshx(1x2)arctantanhx tanh2 (1x2)tanh2xcosh2 1sinh2x(1x2)arctan .(1x2)tanh2xcosh2x (1x2)tanh2xcosh2xg(0)0.g(x)cosh2x12xarctanx,g(0)g(x)2sinh2x2arctanx 1

,g(0) (1x2) 2(1x2g(x)4cosh2x 1x2

(1

4cosh2x 1 14cosh2x

0(当x0时coshx1 1由Taylor公式,对于x0g(x)g(xx30,f(x0,f严格单调递增limf(xlimarctanx故对于x0有arctanx xtanh 2 tanh 证明:当0x时有 tanx sin 证f(xsinxtanxx2f(x)cosxtanxsinxsec2x2xsinxsinxsec2xf(x)cosxsecx2sinxsec2xtanx2(cosxsecx2)2sin2xsecx2(cosxsecxcosx cos

2,x(0,/f(0)f(00,根据Taylorf(x)f(x)x20,sinxtanxx2 tanx(x(0,/ sin 证明下列不等式ex1x,xxx2ln(1x),x0.xx3sinxx,x0.证

ex

ex2

x2

ln(1x)x (1

x2x,xln(1x)xx2 x3xx2,x 3(1 f(xxsinx,f(0)0,f(x1cosx0,仅当x2n时f(x0,故当x0g(x)sinxx

6cosx1,sinxcosx1,sinxg(x) 2 2 严格单调递增g(x)g(0)0,x设xn(1q)(1q2 (1qn),其中常数q[0,1).证明序列xn有极限 q 证xn单调递增.lnxnln(1qq1q1qq

求函数f(xtanx在x4处的三阶Taylor多项式,并由此估计tan(50的值.解f(xsec2x,f(x2xtanx,f(x)4sec2xtan2x2sec4 f()

1,f( 2,f( 4,f( 8 3 f(x)12x 2x x o 8tan(50)tan 12 36 36 336证f(x)ln(1x),f(x)

1

,f(x)

(1 ln(1a) (1 ln(1(1a (1a (1 (1b)bln1(1a (1ab) (1ab)a(1ab)bln1(1a

(1ab)ln(1a abcabc2,abbcca1.0a11b1,1c43 证考虑多项式f(xxa)(xb)(xc)x32x2xabc.f(x)3x24x1(3x1)(x1)0x1x1. 当x1或x1时f(x0,f严格单调递增,当1x11时f(x0,f 如果f(0)f(1)abc0,f实根.如果f(1f44abc0,f f( f(根(见附图).而f实际有根a,b,c.故f(0)f(1)abc0,并且f( f( 27abc考虑到严格单调性,于是在 在(0,,11,)各有一实根正是abc,故结论成立 设函数f(x)的二阶导数f(x)在[ab]上连续且对于每一点x[abf(x)与f同号.证明若有两点cd[ab使f(cf(d0,则f(x0,x[cd证由于f(x)与f(x)同号,(f(x)f(x))f2(xf(xf(x0,g(x)f(x)f(x)单调g(cg(d0故f(x)f(x)0x[cdf2x))2f(xf(x)0x[cdf2xCx[cdf2c0,故f2x0,x[cd],即f(x0x[cd求多项式P3(x2x37x213x9在x1处的Taylor公式 解P(x)6x214x13,P(x)12x14,P(x P3(1)1,P(1)5,P(1)2,P(1) 3P(x)15(x1)(x1)22(x3设Pn(x)是一个n次多项式证明:Pn(x)在任一点x0处的Taylor公式P(x)P(x)P(x) 1P(n)(x n! 若存在一个数a,使P(a)0,P(k)(a)0(k 超过

证(1)Pn(x)是一个n次多项式证明:因为P(x)是一个n次多项式,P(n1x)0,x().故在任一点x P(x)P(x)P(x)(xx) 1P(n)(x)(xx)n n! P(x)P(x)(xx) 1P(n)(x)(xx n!

P(n1)(c)(xx0(n1)!P(x)P(a)P(a)(xa) 1P(n)(a)(xa)nP(a)0(x 故Pn(x)的所有实根都小于

n! 设函数f(x)在(0,)上有二阶导数又知对于一切x0,|f(x|