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/etc/nginx/nginx.conf.求问一道高中导数题 辽宁协作体二模21题 f(x)=1/15x^2-8/15x+1,g(x)=1/3ax^3+1/2bx^2-a^2x(a不等于0)_百度知道
求问一道高中导数题 辽宁协作体二模21题 f(x)=1/15x^2-8/15x+1,g(x)=1/3ax^3+1/2bx^2-a^2x(a不等于0)
则g&#39,3], 即[0, -√a]∪[√a;(x)≥0;6];(x)=ax²(1)=a+b-a²-a)讨论;6)g(x)的值域内∴a²0;2, 0]∵0&-a&(x)&6，g(x)=(1&#47, 即-√a&+bx-a&#178, (1/2, 使得f(x0)=(1/0;1∴f(x)的值域一定不包含于(1/6;6)g(x1)∴实数a要满足的条件是;6)g(x)在[0, g(0)&#47, b=2(2) ∵任取x0∈[0, g&#39, g(0)&#47, 即a≥9时, 即x≤-√a或x≥√a;6)g(x)在区间[0,3]的值域为[g(3)/-a&#178：a) 当a&lt, +∞)上单调递增∵[0;x&lt, g(3)&#47, 总存在x1∈[0，函数f(x)的值域包含于函数(1/6];=a(x&#178, 即[(3a-a²3∴g(1)=a/15)x&#178，解得a=2&#47，若x²0时: 对称轴为x=4∴f(x)在区间[0;6],3]上;)&#47,3]上单调递减值域为[g(3)/6)g(x)在区间(-∞, +∞)∴(1&#47,3]上单调递减, 即[(3a-a²)&#47, (1&#47，值域为[f(3),3];xg'3+b&#47, (1/6;-3a+2≤0∴1≤a≤2若x&#178, 则f(x)的值域包含于(1&#47, 0]∵0&lt,3]包含于[√a;3)ax³=0又极值为-1/0时, 即[(3a-a²0, (3a-a&#178, 0]值域为[g(3)&#47,3]的值域为[g(0)/6)g(x)的值域内f(x)=(1&#47，x²(x)&2≥1;=-1&#47：在区间[0;3联立以上两式;)&#47,3]上单调递减;-(8/2, g(0)/-a≥0, 即[0;6)g(x)在[0;6];∴g'2]∴只需(3a-a²-a&6)g(x)的值域内b) 当a&gt，(1/0, √a)上单调递减∴当√a≥3时, b=-2/9或a=-1;)/(1)=0又g'2-a²√a;15)x+1;-a²6, 则g&#39, g(x)在区间[0;(x)=ax²)/1∴f(x)的值域一定不包含于(1&#47, f(0)];3;6)g(x)在区间(-√a, 1]b=0时(1) ∵g(x)在x=1处取极值∴g&#39
采纳率：70%
来自团队：
x&lt，对于任意0&lt.5]间有一个极值点;=0;3-a^2x,3]之间是单调递降根据题意,x0;=3;2-a^2=-1/3 or a=-1;=6g(x)=ax(x^2-3)/=3;a&(1)=0,a+b-a^2=0,x=3^0；f(x)在[0;6f(x)=(x-3)(x-5)&#47；g(x)在[0,0&lt,3^0;2-a^2=-1&#47,f(x)在x=3和x=5除有两零点;b=-5/3b=a^2-a,x在[0;3=0,b=0;15-8x/3+(a^2-a)&#47，使f(x0)=g(x1)/g(1)=a&#47,3],a=2/3+b&#47g(x)=ax^3&#47；x&lt,3]的值域[k1;3+bx^2&#47,f(x0)=g(x1)&#47,(3a-2)(a+1)/9 or b=3f(x)=x^2&#47,3]值域为[1,k1&2-a^2x(a不等于0)g'3^0;6=0;=0,k2&gt.5处有三个零点，总能找到0&lt,f(x)&gt,g'3,x1:[0.5,g(x)=ax^3/=x0&6-1/2+a/6成立，g(x)在x=-3^0,得x^2=a,0];15.5,a/3;=x1&lt,令g'=3,a^2&#47.5,0&(x)=ax^2+bx-a^2;(X)=ax^2-a^2=0;15+1,k2],x=0,x=a^0,则g(x)在[0
第二问呢？？~~
由于3&a&0,g(x=a^0.5)=a^2.5/3-a^2..5=-2a^3/3&0所以g(x&3^0.5)&0因此g(3)&=k2&=6,a*3^3/3-a^2*3=9a-3a^2&=6,a^2-3a+2&=0,(a-1)(a-2)&=0,1&=a&=2
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