The known function f x a 3 x 3 a 1 2 x 2 x b, where a, b belong to R

1) Derivative of a 3x3-(a 1) 2x2 x b, f(x)' = ax 2-(a 1)x 1

Because the slope of the equation y=5x-4 at the tangent is 5 and the point p(2,f(2)) i.e. f(x)'=5 when x=2

Substituting the two into f(x)'=ax 2-(a 1) 1 has 5=4a-2(a 1) and the solution is a=3

So f(x)=x 3-x 2 x b

When x=2, y=5x-4=6 is substituted for f(x)=x 3-x 2 x b, and 6=8-4 2 b is obtained, so b=0

So f(x)=x3-x2 x

2) From f(x)'=ax 2-(a 1)x 1 so that f(x)'=0 has ax 2-(a 1) x 1=0 δ=(a 1) 2-4a

So x1=[(a 1) a-1) 2] (2a) x2=[(a 1)- a-1) 2] (2a).

When 0x2=1

So when x>=1 a or x=<1, f(x)=x 3-x 2 x increases monotonically;When 11, the same can be found when 1=1 a or x=<1 f(x)=x 3-x 2 x monotonically decreasing.

The topic is not clear, ** some spaces, and the operation is in line with the writing perfection point.

(1) f(x)=x -ax+b decreases monotonically in the interval (- 1), a 2>=1, a>=2

2) There is a real number a, such that when x [0,b], 2 x -ax+b 6 is constant, x = 0 when 2 b 6, x = b 2 b 2-ab + b 6, x 2-ax + 6 = (x-a 2) 2+6-a 2 4, from 6-a 2 4 > = 2, a 2,10, let the function f(x) = x -ax + b, a, b belong to r

Knowing that f(x) decreases monotonically over the interval (negative infinity, 1), find the range of values of a.

1) Derivative of a 3x3-(a 1) 2x2 x b, f(x)' = ax 2-(a 1)x 1

Because the slope of the equation y=5x-4 at the tangent is 5 and the point p(2,f(2)) i.e. f(x)'=5 when x=2

Substituting the two into f(x)'=ax 2-(a+1) has 5=4a-2(a+1) and the solution is a=3

So f(x)=x 3-x 2 x b

When x=2, y=5x-4=6 is substituted for f(x)=x 3-x 2 x b, and 6=8-4 2 b is obtained, so b=0

So f(x)=x3-x2 x

2) From f(x)'=ax 2-(a+1) so that f(x)'=0 has ax 2-(a+1)x+1=0 δ=a+1) 2-4a

So x1=[(a 1) a-1) 2] (2a) x2=[(a 1)- a-1) 2] (2a).

When 0x2=1

So when x>=1 a or x=<1, f(x)=x 3-x 2 x increases monotonically;When 11, the same can be found when 1=1 a or x=<1 f(x)=x 3-x 2 x monotonically decreasing.

In summary,..Don't write it anymore, hehe, finally the code is over, and the landlord will give it.

1 From the condition of the problem, f(1)=2, the value of the derivative function of f(x) at x=1 is 2, so that 1 3+a+b=2, 1+2a+b=2, and a=-2 3, b=7 3

2 From the meaning of the question, the derivative function of f(x) is 0 in the interval (1,2), so that the equation x 2+2ax+b=0 has two roots in (1,2), which is equivalent to 1<-2a (2*1)<2, the discriminant formula of the equation is greater than 0, and f(1)>0, f(2)>0, the solution is -2-2a-1, so 0<-2a-1+a

,1、f'(x)=x 2ax b, then f'(0)=b=1 and f'(2)=4 4a b=1, the solution is a=b='(x)=x²－ax＋(a＋2)【

Quadratic functions.! ] is greater than or equal to 0 on (0,1), then:

The axis of symmetry is 0 and f'(0) 0 or axis of symmetry 1 and f'(1) 0 or 0, solution.

,1、f'(x)=x 2ax b, then f'(0)=b=1 and f'(2)=4 4a b=1, the solution is a=b='(x)=x ax (a 2)【Quadratic Function!! ] is greater than or equal to 0 on (0,1), then: axis of symmetry 0 and f'(0) 0 or axis of symmetry 1 and f'(1) 0 or 0, solution.

'(x)=x^2-2ax+b

f '(0)=f '(2)=1 to obtain b=1, a=1f(x)=1 3x 3-x 2+x

x)=x^2-2ax+b

i.e. f'(x) in (0.)., 1) on the minimum value of 0

Discuss a<=0,00, a>-2 or -1

1.f can be known from the function'(x)=-1 x 4-2ax+b, and then substituting 0 and 2, solving the equation yields the values of a and b. The second question can be answered by the fact that the derivative is greater than zero. a is less than 1

1 From the condition of the problem, f(1)=2, the value of the derivative function of f(x) at x=1 is 2, so that 1 3+a+b=2, 1+2a+b=2, and a=-2 3, b=7 3

2 From the meaning of the question, the derivative function of f(x) is 0 in the interval (1,2), so that the equation x 2+2ax+b=0 has two roots in (1,2), which is equivalent to 1<-2a (2*1)<2, the discriminant formula of the equation is greater than 0, and f(1)>0, f(2)>0, the solution is -2-2a-1, so 0<-2a-1+a

f'(x)=(x²+ax-2a²+3a+2x+a)e^x=[x²+(a+2)x-2a(a-2)]e^x=(x+2a)(x+2-a)e^x

by f'(x)=0, we get x1=-2a, x2=a-2 because a≠2 3, then x1≠x2

Therefore, x1 and x2 are extreme points, f(x1=3ae (-2a), f(x2)=(-3a+4)e (a-2).

1) When a>2 3, x2 > x1

The monotonic increase interval is: xx2; The monotonic reduction interval is (x1, x2) and the maximum is f(x1)=3ae (-2a).

The minimum is f(x2)=(-3a+4)e(a-2)2) when a<2 3, x1>x2

The monotonic increase interval is: xx1; The monotonic reduction interval is (x2, x1), the maximum is f(x2)=(-3a+4)e (a-2), and the minimum is f(x1)=3ae (-2a).

Derivative f'(x)=3x+2(1-a)x-a(a+2)let f'(x)=0,(x-a)(3x-(a+2))=0x=a or x=(a+2)3

Not monotonous on (-1,1).

f'(x) There is a zero point on x (-1,1).

1 (a-2) 3 1 or -1 a 1 i.e. -1 a 5 or -1 a 1

i.e. -1 a 5

There are different numbers around zero o'clock.

Discriminant formula =4(1-a) +12a(a+2)=16a -16a+4=4(2a-1) 0

i.e. a≠1 2

The value of a can be (-1,1 2) (1,2,5).

b^2 - 4ac != 0 in -1 to 1 There is a solution, an equation is solved, and the image method is fine.