tanx+2
钱旺儿2822证明不等式tanx+2sinx>3x(0<x<二分之派).要步骤啊! -
庄别录17368603003 ______ tanx+2sinx>3x(0设f(x)=tanx+2sinx-3x f(0)=0 f'(x)=sec²x+2cosx-3 f'(0)=0 f"(x)=2sec²x tanx -2 sinx=2sinx(sec²xsecx-1)>0 secx是cosx的倒数,在(0所以,当(0有f"(x)>0,==>f'(x) 单增 而f'(0)=0 ===>f'(x)>0 ==>f(x)单增 而f(0)=0 ===>f(x)>0 即tanx+2sinx>3x
钱旺儿2822求函数y=sec^2x+tanx+2的最值x属于[ - π/6, π/4] -
庄别录17368603003 ______ y=sec²x+tanx+2 =sec²x-1+1+tanx+2 =tan²x+tanx+3 =(tanx+1/2)²=11/4 x∈[-π/6,π/4] tanx∈[-根号下3/3...
钱旺儿2822求函数y=(tanx)的平方 - 2乘tanx+2的值域和单调区间 -
庄别录17368603003 ______[答案] 求函数y=(tanx)^2-2*tanx+2的值域和单调区间 y=(tanx^2-2tanx+2) 令tanx=k 属于R y=[(k-1)^2+1] 令t=(k-1)^2>=0 y=(t+1) 值域是y>=1 根据复合函数可知 tanx>1 单调增 tanx
钱旺儿2822已知 - π/3≤x≤π/4,f(x)=tan^2x+2tanx+2,求f(x)的最值及对应的x值 要具体的解题步骤 -
庄别录17368603003 ______ f(x)=(tanx+1)^2+1 故:x=-PI/4时tanx =-1 此时f(x)最小 为1 tan x在相应的x范围内递增 所以 -根号3 考虑两个边界值 tanx=-根号3 时 f(x)=5-2倍根号3 tan x=1时f(x)=5 故:f(x)最大值是5, 相应x=pi/4 综上所述:x=-pi/4时tanx =-1 此时f(x)最小 为1 x=-pi/4时tanx=1 此时f(x)最大 为5
钱旺儿2822已知 - π/3≤x≤π/4,f(x)=tanx+2tanx+2,求f(x)的最值,及相应的x值拜托各位了 3Q -
庄别录17368603003 ______ 解:令t=tanx, 由于-π/3≤x≤π/4,故-√3≤t≤1,f(x)=t+2t+2=(t+1)+1, 所以,当t=-1时,即x=-π/4时,f(x)取得最小值1; 当t=1时,即x=π/4时,f(x)取得最大值5.
钱旺儿2822若x∈[ - π/3,π/4],求函数y=1/cos^2x+2tanx+1的最知及相应的x值 -
庄别录17368603003 ______[答案] y=sec²x+2tanx+1=tan²x+2tanx+2=(tanx+1)²+1 x∈[-π/3,π/4],则tanx∈[-√3,1] 当tanx=-1,即x=-π/4时,y有最小值1 当tanx=1,即x=π/4时,y有最大值5
钱旺儿28222tanx+1=? -
庄别录17368603003 ______[答案] tanx^2+2tanx+1=(tanx+1)^2 tanx^2+1=secx^2=1/cosx^2 sinx^2+cosx^2=1
钱旺儿2822已知tanx=2,则sin2x+2cos2x= -
庄别录17368603003 ______[答案] 1、(sin2x+2cos2x)/(sin²x+cos²x) =(2sinxcosx+2cos²x-2sin²x)/(sin²x+cos²x) 分子分母同时除以cos²x得 =[2tanx+2-2(tanx)²]/[(tan)²+1] 带入 tanx=2得 =(4+2-8)/(4+1) =-2/5 2、tanx=2(第一象限) tan2x=4/1-4=-4/3(第二象限) sin2x=4/...
钱旺儿2822matlab求解导数y=ln((2tanx+1)/(tanx+2)),y=sin(e^(x^2+3x - 2)) -
庄别录17368603003 ______[答案] y1 = 'log((2*tan(x)+1)/(tan(x)+2))' %log在matlab中求自然对数 y11=diff(y1) %求导 simple(y11) %化简 y2 = 'sin(e^(x^2+3*x-2))' y21=diff(y2) simple(y21)