It is known that f x ax 2 b if a 2, b 3, finds the tangent equation of Q 0, 1

It is known that f(x)=ax2+bIf a=2, b=3f(x)=2x2+3

f'(x)=4x

Let the tangent point of q(0,-1) be (m,n).

then k=f'(x)|(x=m)=4m

k=(n+1)/m k=4m

n+1=4m^2

The tangent point is on the curve n=2m 2+3

4m^2-1=2m^2+3

2m^2=4

m 2=21) m= 2 k=4 2 tangent equation y=4 2x-12) m=- 2 k=-4 2 tangent equation y=-4 2x-1

Substituting a=2,b=3 into f(x)=ax 2+b.

f(x)=2x^2+3

f'(x)=4x

Let the tangent point of q(0,-1) be (x0,y0), then k=f'(x)|(x=x0)=4x0

k=(y0+1)/x0 ,k=4x0

y0+1=4x0^2

The tangent is on the curve.

y0=2x0^2+3

4x0^2-1=2x0^2+3

2x0^2=4

x0^2=2

The solution yields x0= 2 or x0=- 2

k=4 2 or k=-4 2

The tangent equation over q(0,-1) y=4 2x-1 or y=-4 2x-1

f(x) =3ax 2+2bx-3 is f(1) =0 from the problem, i.e. 3a+2b-3=0 a=1

f(1)=-2, i.e., a+b-3=-2 bipolar b=0 is f(x)=x 3-3x

f(x) =3x 2-3 Therefore, the slope of the tangent of (x,f(x)) is f(x) =3x 2-3

If the tangent of the point (x,f(x)) is y= f(x) x+b, then the point (x,f(x)) must be passed and the point (x,f(x)) must be substituted into x 3-3x=(3x 2-3)x+b

b=-2x 3, i.e., the tangent equation is y= f(x) x-2x 3

Substituting the Zheng point m(2,m) into the tangent y= f(x) x-2x 3 gives m=2(3x 2-3 )-2x 3 to 2x 3-6x 2+6+m=0

From the problem, it can be seen that the equation 2x 3-6x 2+6+m=0 has three different solutions, which can be understood as its minimum value is less than 0 and the maximum value is greater than 0

Let g(x)=2x 3-6x 2+6+m g(x) =6x 2-12x=0 to obtain x=0 or 2, and when it is judged that g(x) is at x=0, g(x) obtains the maximum value of 6+m. When x=2, g(x) obtains a minimum value of m-2

6+m>0

m-2 "Dust line 0 gets -6, so the value range of m is -6

It is easy to know that a=1, b=0, set the tangent coordinates as (x0, y0), and the tangent equation is y-yo=(3x0 2-3)(x-x0), and bring x=2, y=m, y0=x0 3-3x0 into the m=-2xo 3+6x0 2-6 Lingxvertical f(x)=-2x 3+6x 2-6 f'(x)=-6x 2+12x Apparently f(x) is monotonically decreasing at (negative infinity, 0), Sun Changshi (2, positive infinity) is monotonically decreasing, and increasing monotonically at (then limb 0, 2). The three tangents of f(x) can be made, which means that y=m has 3 intersections with f(x), f(0).

1) f(x)=ax+b(x-1)-a then f'(x)=a-b/(x-1)^2

At x=3 there are: and mu f'(3)=a-b/4

In addition, at x=3, the cutting sock line equation is (2a-1), x-2y+3=0, then its slope is (2a-1) 2

So f'(3)=a-b/4 =（2a-1）/2

The solution gives b=2 so: g(x)=f(x+1)=a(x+1)+2 x-a=ax+2 x

g'(x)=a-2/x^2

Tangent at any point on the curve g(x): y=(a-1 x0 2)(x-x0)-y0

Then the tangent intersects with the straight Hengsen line x=0 and the straight line y=ax at (0,2 x0+y0-ax0), (x0-ax0 3 2+y0x0 2 2,ax0-a 2x0 3 2+ay0x0 2 2), respectively

Then the area of the triangle enclosed by the tangent of any point on the curve g(x) and the line x=0 and y=ax is.

s=(2/x0+y0-ax0)(x0-ax0^3/2+y0x0^2/2)/2

y0=ax0+2 x0

The solution yields s=4 as a fixed value.

Answer: This is a national college entrance examination question. It seems to be in 2004. (to be checked).

I'll give you an answer.

Solution: The derivative f'(x)=3x +2 follows f'(x)=3x +2ax+b

The function f(x)=x 3+ax 2+bx+2 and the straight line 4x-y+5=0 are tangent at the point p(-1,1), f'(-1)=3(-1) +2a(-1)+b, f(-1)=(-1) 3+a(-1) 2+b(-1)+2, and the simultaneous simplification obtains: a-b=0, b-2a=-2, and the solution: a=b=2,

(1).To cut to the point, it can give us two pieces of information.

First, it's on the curve.

Second, the slope of this point is equal to the slope of the straight line.

The slope is the derivative, then the derivative of the curve is required.

f (x) = 3x 2 + 2 ax + b, linearized to y = 4x + 5, the slope is 4

can be obtained. The derivative of x = -1 points is 4, and f (-1) = 3-2a + b = 4

f(-1)=-1+a-b+2=1.

Synopid a=-1, b=-1

2）。f(x）=x^3-x^2-x+2.

f(x) mx 2-x-2, constant formation, shifting, finishing x 3-(1+m) x 2+4 0 constant formation.

Construct a new function, g(x)=x 3-(1+m)x 2+4, and find its monotonicity.

g`(x)=3x^2-2(1+m)。

Discussion , m 0, the derivative is all 0, i.e. on x [1,2] is an increasing function.

From the constant formation, the minimum value of 0, i.e., g(1) 0, and the solution is m 4Take the intersection with m 0, get m 0，m≥0,g`(x)=3x^2-2(1+m)=3*(x^2-2/3*(1+m))=3*[x+√(2/3*(1+m))]x-√(2/3*(1+m))]

So the subtraction interval is [- 2 3*(1+m)),2 3*(1+m))].

Then draw a diagram and discuss the relationship between [1,2] and this interval.

It's too ink, why don't you try it yourself.

If you don't wait for me for a day, I'll get back to you tomorrow night.

Substituting x=1 into the tangent equation yields y=5 2, that is, bringing (1,5 2) into the ordinate at x=1 yields: 1+b+c=5 2,b+c=3 2 The derivative of f(x) yields: 3x 2+2bx+c

The tangent slope k=3, then x=1 is brought into the above equation to obtain 3, that is: 3+2b+c=3 simultaneous solution; b=-3/2,c=3

Solution: From the problem setting, it can be known that f(x).'=3x^2+2bx+cf(1)'=3+2b+c=3

f(1)=1+b+c=

Solving the above two equations yields: b=, c=3

f'(x)=x 0 5-2ax+(a 0 5-1) because at x=1 the slope of the tangent line of the source of regret is split-1, so f'(1)=1-2a+a�0�5-1=-1

The tangent equation of a=1 can be written as y-f(1)=-x-1)x+y-1-f(1)=0

f(1)=2

So (1 3)-a+a 0 5-1+b=2b=-8 3

Tell you how to find the analytic and monotonic interval method, the steps of solving this kind of problem are basically like this, usually do more such problems!

Finding the derivative of the function f(x) first gives f(x).'=a+b/x^2-2/x

Since the tangent equation for f(x) at x=1 is x+4y-2=0, then y=(-1 4)x+1 2

Substituting x=1 into f(x).'Get a+b-2=-1 4 (1).

Substituting x=1 into f(x) gives a-b=-1 4+1 2 (2).

From equations (1) and (2), the values of a,b can be obtained.

Then substituting the values of a and b into f(x) to get the analytic formula.

Then substitute the values of a and b into f(x).', let f(x).'is equal to 0, and the solution of the equation can solve the monotonic interval of f(x).

The second problem is to bring the f(x) equation into an equation, and then find the value of x that meets the requirements in the interval [1 4,2] (also composed of intervals), and then sort out the function y=g(t)=t 2+t-2 into the form of y=(t+1 2) 2-9 4, and then substitute the obtained x value to obtain the maximum value.