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设F(x)=x3+Ax2+Bx+1的导数F′(x)满足F′(1)=2A...

(I)∵f(x)=x3+ax2+bx+1∴f'(x)=3x2+2ax+b.令x=1,得f'(1)=3+2a+b=2a,解得b=-3令x=2,得f'(2)=12+4a+b=-b,因此12+4a+b=-b,解得a=-32,因此f(x)=x3-32x2-3x+1∴f(1)=-52,又∵f'(1)=2×(-32)=-3,故曲线在点(1,f(1))处的切...

(I)因为f(x)=x3+ax2+bx+1,所以f'(x)=3x2+2ax+b.…..(2分)令x=1得f'(1)=3+2a+b.由已知f'(1)=2a,所以3+2a+b=2a.解得b=-3.….(4分)又令x=2得f'(2)=12+4a+b.由已知f'(2)=-b,所以12+4a+b=-b,解得a=?32.…..(6分)所以f(x...

答:递增区间为(-∞,-1)U(3,+∞),递减区间为(-1,3) f(x) = x³ + ax² + bx + 1 f'(x) = 3x² + 2ax + b f'(1) = 2a - 6 => 3 + 2a + b = 2a - 6 => b = -9 f'(2) = - b - 18 => 12 + 4a + b = - b - 18 => a = -3 f(x) = x³...

解答:解:f′(x)=3x2+2ax+b由f′(?1)=3?2a+b≤2f′(1)=3+2a+b≤2 得 2a?b?1≥02a+b+1≤0不等式组确定的平面区域如图阴影部分所示:由2a?b?1=02a+b=1=0 得a=0b=?1,∴Q点的坐标为(0,-1).设z=ba?1,则z表示平面区域内的点(a,b)与点P(1...

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f'(x) = 3x^2+2ax+b 因为f'(1)=2a,f'(2)=-b,把当 x=1和x=2分别代入上式则 3 + 2a +b =2a , 12+4a + b = -b,求出 b = -3, a = - 3/2 f(x)=x^3 - 3/2x^2 - 3x +1 (1) 曲线y=f(x)在点(1,f(1))处的切线斜率为f'(1)=2a=-3 f(1) = -5/2 y + 5/2 = -...

∵f(x)=x3-(2a+2)x2+bx+c,∴f′(x)=3x2-4(a+1)x2+b,∵曲线y=f(x)在与x轴交点处的切线为y=x-1,∴曲线y=f(x)在与x轴交点为(1,0),则f(1)=-1-2a+b+c=0,f′(1)=-1-4a+b=1,①∵函数f(x)的导数y=f′(x)的图象关于直线x=2对称,∴2(a...

∵f′(x)=3x2-2(a+1)x+a-2,∴f(x)=x3-(a+1)x2+(a-2)x+m,∴f(0)=m=2a,∴f(x)=x3-(a+1)x2+(a-2)x+2a=x3-x2-2x-ax2+ax+2a(a-2)x+2a=x(x2-x-2)-a(x2-x-2)=(x-a)(x2-x-2)=(x-a)(x-2)(x+1)<0∵a>2,∴不等式f(x)<0...

(Ⅰ)设g(x)=f′(x)=2+1?2ax+a=2x+1x+a,则g′(x)=2a?1(x+a)2,当a≥12时,函数y=f′(x)在区间(-∞,-a]、(-a,+∞)上单调递增,当a<12时,函数y=f′(x)在区间(-∞,-a]、(-a,+∞)上单调递减,∴函数y=f′(x)的单调区间是(-∞,-a]、...

(1)依题意,得f′(x)=3x2-3a2,f″(x)=6x,令f″(x)=0,得x=0,又f(0)=2a,所以g(x)的“拐点”为(0,2a)(2)φ(x)=a(x+1)3+b(x+1)+3 (a≠0)(也可以据此给定a,b的值,写出几个关于x的函数);(3)g′(x)=3(x2-a2)因此a≥1,...

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