已知函数F(X)=X^3+AX^2+BX+2与直线4X-Y+5=0相切于X=1处,若X>0时,不等式F(X)≥MX^2-
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/25 09:39:23
已知函数F(X)=X^3+AX^2+BX+2与直线4X-Y+5=0相切于X=1处,若X>0时,不等式F(X)≥MX^2-2X+2恒成立,求实数M的取值范围?
F(X) = X^3+AX^2+BX+2 与直线 4X-Y+5 =0 相切于X=1处
F(1) = 1 + A + B + 2 = 3 + A + B,
4 - Y + 5 = 0,Y = 9,
9 = Y = F(1) = 3 + A + B,
A + B = 6 ...(1)
F'(X) = 3X^2 + 2AX + B,
F'(1) = 3 + 2A + B,
4X - Y + 5 = 0,
Y = 4X + 5.斜率为4,
因此,
4 = F'(1) = 3 + 2A + B,
2A + B = 1 ...(2)
由(1),(2)解得,
A = -5,
B = 11.
所以,
F(X) = X^3 - 5X^2 + 11X + 2.
令G(X) = F(X) - MX^2 + 2X - 2 = X^3 - 5X^2 + 11X + 2 - MX^2 + 2X -2
= X^3 -(5+M)X^2 + 13X
= X[X^2 -(5+M)X + 13].X>0
则当X>0,G(X) > 0时,总有,
X^2 - (5+M)X + 13 > 0
(5+M)^2 - 4*13 < 0
25 + 10M + M^2 - 52 < 0,
M^2 + 10M - 27 < 0,
Delta = 10^2 + 4*27 = 100 + 108 = 208 = 4^2*13,
[-10 -Delta^(1/2)]/2 < M < [-10 + Delta^(1/2)]/2,
[-10 -4(13)^(1/2)]/2 = -5-2(13)^(1/2) < M < [-10 + 4(13)^(1/2)]/2 = -5 + 2(13)^(1/2),
实数M的取值范围,
-5-2(13)^(1/2) < M < -5 + 2(13)^(1/2).
F(1) = 1 + A + B + 2 = 3 + A + B,
4 - Y + 5 = 0,Y = 9,
9 = Y = F(1) = 3 + A + B,
A + B = 6 ...(1)
F'(X) = 3X^2 + 2AX + B,
F'(1) = 3 + 2A + B,
4X - Y + 5 = 0,
Y = 4X + 5.斜率为4,
因此,
4 = F'(1) = 3 + 2A + B,
2A + B = 1 ...(2)
由(1),(2)解得,
A = -5,
B = 11.
所以,
F(X) = X^3 - 5X^2 + 11X + 2.
令G(X) = F(X) - MX^2 + 2X - 2 = X^3 - 5X^2 + 11X + 2 - MX^2 + 2X -2
= X^3 -(5+M)X^2 + 13X
= X[X^2 -(5+M)X + 13].X>0
则当X>0,G(X) > 0时,总有,
X^2 - (5+M)X + 13 > 0
(5+M)^2 - 4*13 < 0
25 + 10M + M^2 - 52 < 0,
M^2 + 10M - 27 < 0,
Delta = 10^2 + 4*27 = 100 + 108 = 208 = 4^2*13,
[-10 -Delta^(1/2)]/2 < M < [-10 + Delta^(1/2)]/2,
[-10 -4(13)^(1/2)]/2 = -5-2(13)^(1/2) < M < [-10 + 4(13)^(1/2)]/2 = -5 + 2(13)^(1/2),
实数M的取值范围,
-5-2(13)^(1/2) < M < -5 + 2(13)^(1/2).
已知函数f(x)=x^3+ax^2+bx的图像与直线9x-y+8=0相切于(-1,-1)
已知二次函数f(x)=ax^2+bx,f(x+1)为偶函数,函数f(x)的图象与直线y=x相切
已知二次函数f(x)=ax^2+bx,f(x+1)为偶函数,函数f(x)的图象与直线y=x相切.
已知二次函数f(x)=ax^2+bx,f(x+1)为偶,函数f(x)的图像与直线y=x相切.
已知函数f(x)=ax^3+bx^2+c(x属于R)的图像与直线L:15x-y+10=0相切于点(-1,-5),且函数f
有关极限的,已知函数f(x)=ax^2+bx+c,满足f(0)=3,且直线y=5x+1与图像相切于点(2,11)(1)求
已知函数f(x)=x^3+ax^2+bx+c在x=-2处取得极值,并且他的图像与直线y=-3x+3在点(1,0)处相切,
已知函数f(x)=x^3+ax^2+bx+c在x=2处取得极值,并且它的图像与直线y=-3x+3在点(1,0)处相切,求
已知函数f(x)=x^3+ax^2+bx+c在x=2处取得极值,并且它的图像与直线y=-3x+3在点(1,0)处相切 求
已知二次函数f(x)=ax^2+bx,f(x+1)为偶函数,函数f(x)的图像与直线y=x相切.(1)求f(x)的解析式
已知a>0,函数f(x)=ax^2-x,g(x)=ln(ax).(1)若直线y=kx-1与函数f(x),g(x)相切与同
已知函数f(x)=x^2+ax,x≥0,bx^2-3x,x<0为奇函则不等式f(x)