A high school comprehensive problem of functions and derivatives

f(x)=kx+lnx, when k=0 is, is there an unequal positive number a, b satisfying [f(a)-f(b)] (a-b)=f '[a+b) 2]?

Solution: When k=0, f(x)=lnx, f(x)=1 x, lna-lnb=ln(a b)=(a-b)[2 (a+b)]=2(a-b) (a+b).

Let a b = m, a = mb, ln(a b) = ln(mb b) = lnm = 2 (mb-b ) (mb + b) = 2 (m-1) (m+1).

Therefore, as long as m is found, so that the equation lnm=2(m-1) (m+1) holds, (m≠1), the original proposition holds.

Let y =lnx,y =2(x-1) (x+1)=[2(x+1)-4] (x+1)=2-4 (x+1).

It is not difficult to see from the graph: there is only one intersection point (1,0) between the logarithmic curve y =lnx and the curve y =2-4 (x+1), and if x=m=1, then there is a=b, which is inconsistent with the meaning of the topic, so there is no unequal positive number a,b satisfies [f(a)-f(b)] (a-b)=f [a+b) 2].

When k=0, f(x)=lnx, f(x).'=1/x,[f(a)-f(b)]/(a-b)=[ln(a/b)]/(a-b),f'[(a+b) 2]=2 (a+b), then only :ln(a b)=2, which can be done, just a=eb

Step 1: Find B

Knowing f(x), we can find f'(x), i.e. f'(x)=x^2-(a+1)x+b

And from the meaning of the question, it can be seen that (0,0) satisfies f'(x)=x 2-(a+1)x+b, so substitute the point (0,0) into f'(x)=x 2-(a+1)x+b can be obtained: b=0

1) Because a=1, b=0 so f'(x)=x^2-(a+1)x+b=x^2-2x

When x=3, k=f'(3)=3 (k is used to represent the slope of the function f(x) at x=3).

Because f(x) = (1 3) x 3-x 2+1 , so f(3) = 1

Therefore, the tangent equation for the image of the function f(x) at x=3 is y=3x+b, (the key is to find b).

From the meaning of the question, we can know that the point (3,1) is on the straight line y=3x+b, so substituting the point, we can find b= -8

So the tangent equation for the image of the function f(x) at x=3 is y=3x-8

2) From the meaning of the title, we can see that x 2-(a+1)x=-9, that is, the existence equation x 2-(a+1)x +9=0

Problem 2 becomes: If the equation x 2 - (a + 1) x + 9 = 0 has a solution at x<0, find the maximum value of a.

This should be the knowledge of the second year of high school, you should think for yourself!

(3) Because f'(x)=x 2-(a+1)x let f'(x)=0 gives x=0 or x=a+1

Because A+1>0

f(x) increases on (negative infinity, 0] and [a+1, positive infinity) and subtracts on (0,a+1).

f(0) max = a>0

f(a+1)=(-a 3-3a 2+3a-1) 6 pairs (-a 3-3a 2+3a-1) 6 derivation: when a=2 (1 2)-1, f(a+1) maximum=18 (1 2)-7<0, then f(x) has three zeros.

Derived from the question:

f(x)'=(3-a) 3*x(2-a)+x+b due to f(x).'Crossing the origin (0,0).

0=(3-a)/3+b→a-3b=3

1) When a=1, b=-2 3

f(3)'=

f(1)=3

by f(2t-1) 2f(t)-3

f(2t-1)-f(t) f(t)-f(1) i.e. the function f(x) increases monotonically at [1,+.

For any x 1

f'(x)=2x+2+a/x

2x +2x+a) x 0 constant established.

a≥-2x²-2x

-2x -2x 0 at x 1

Get a 0

<> second question a 15 3? Not very sure,

g(x)=x³-2x²+x+λx²-λ

x³+(2)x²+x-λ

g'(x)=3x²+2(λ-2)x+1

The increment function is x>0, g'(x) Evergrande is at 0

If g'The discriminant formula for (x) is less than 0

then g'(x) is greater than 0 on r

then 4( -2) -12<0

If the discriminant is greater than or equal to 0

then <=- 3+2, >= 3+2

At x>0, g'(x) Evergrande is at 0

It must be on the right side of the axis of symmetry.

That is, the axis of symmetry x=-(2) 2<=0

In this case, it's an increment function.

g'(0)=1>0

So x>0, g'(x) >0 is established.

So > = 3+2

In summary, >-3+2

Original formula = (m -mn-5mn + 5n) -6 (m -3mn + 2mn-6n).

m²-6mn+5n²)-6(m²-mn-6n²)=m²-6mn+5n²-6m²+6mn+36n²=-5m²+41n²

Left and right identities. So the coefficients of terms with the same number of x degrees should be equal.

i.e. the x-coefficients are equal, where a=a

Then the coefficients and constant terms of x are equal, respectively.

So we get this system of binary equations.

According to the derivative definition, tangent means that the slope is the same.

l slope = kf(x) = lnx,f'(x)=1 x, the slope is f at x=1'(1)=1

The coordinates at x=1 are (1,0).

The equation for a straight line is y=1(x-1)=x-1

g'(x)=x+m=1

Let the line be tangent to g(x) at the points (a, b).

There should be b=a-1

a+m=1a^2/2+ma+7/2=b

Solution a 2 = 9

a=+ -3

m=-2 or 4

If you don't understand, you can ask me.

x)=x^2+bx+c

h'(x)=2x+b, then h(x)-h'(x)=x 2+(b-2)x+c-b.

then the discriminant formula: (b-2) 2-4(c-b)<=(less than or equal to) 0, then 4c>=b 2+4>=4|b|

then c>=|b|

h(b)=2b^2+c

Then h(c)-h(b)=c 2+bc-2b 2=(c-b)(c+2b) is established by the constant condition.

c-b)(c+2b)<=m(c2-b 2)constant hold, then m>=(c-b)(c+2b) (c 2-b 2) approximate: m>=(c+2b) (c+b).

Separation coefficient: m>=1+b (c+b)=1+1 c b+1 (fraction up and down divided by b) because: c>|b|

then 1+1 c b+1 is less than.

then the minimum value of m is .

According to the intention of the questioner, the answer on the first floor is well written, but the first question of this question is a mistake. Since f(x) has two different extreme points, then b 2-4c > 0, but from the first question condition, 4c >=b 2+4 then b 2>4c>=b 2+4, which is impossible

What I said upstairs makes sense, I think the condition that there are two extreme points should be removed.

This is a college entrance examination question, in recent years, it is not difficult but it must be rigorous, otherwise you can't do the second question, and then think about it yourself.

The lengths on both sides of the rectangle are x and y, respectively

According to the title, x+ y=8, i.e. x=8-y

The area enclosed by the cylinder is s= xy 2 and x=8-y is brought into the area with s= (8y 2-y 3).

Then s'= y(16-3y) When s'=0 there is 16-3y=0 or y=0 (rounded).

So there is y=16 3 to bring in x+y=8 and there is x =8 3

1) f(x)=[(1+x)/(1-x)]e^(-ax)f'(x)=[2/(1-x)^2-a(1+x)/(1-x)]e^(-ax)

e^(-ax)[2/(1-x)^2-a(1-x^2)]/(1-x)^2

e^(-ax)[2-a(1-x^2)]/(1-x)^2e^(-ax)*a[x^2+(2/a-1)]/(1-x)^2 ..a>0

0=0 f(x) increases monotonically at x<1 and x>1.

a>=2, x>1 f'(x)>0 f(x) monotonically increases.

1-2 a) 0 f(x) monotonically increased.

1-2/a)1

0 is obtained from (1).