Knowing the function f x ln x 1 ax, if x 0, f x takes the maximum value and finds the value of the r

Are you a high school student or a college student? Because the methods used are different, can you use "derivatives" at your current learning stage?

If you are a high school student, it is recommended to use the ** method, and the process is as follows:

When x=0, f(x)=0

The first step is to draw an image of ln(x+1), which you will, right? The image passes through the origin, and the shape is the same as that of LNX, which is equivalent to the latter translating to the origin at the intersection point (1,0) of the x-axis.

In the second step, the image of y=ax represents some straight lines, where a is the slope of the straight line, but a is to be determined.

In the third step, the logarithmic function y=ln(x+1) and the straight line y=x are tangent at the origin (0,0). Common sense: There is always x>=ln(x+1), and the only place with the equal sign is x=0This truth can be seen by drawing a picture. At this point, a=1.

In the fourth step, it can be seen that if a>=0, then f(x)=0 cannot be a maximum, because with x>0, f(x) >0;

In the -10 segment, it is possible that f(x)=ln(x+1)+ax>0 (because the line y=-ax, which is symmetrical with y=ax with respect to the y-axis, intersects ln(x+1), and the drawing is visible), so that x=0 is not a maximum point.

If a<-1 is at -10 (because this line y=-ax intersects ln(x+1), then x=0 is not a maximum point.

If a=-1, whether it is at -10, there must be f(x)=ln(x+1)+ax<0, so f(x)=0 is the maximum value when x=0.

To sum up, the real number a=-1.

If you are a college student, use the "derivative" to solve (the solution in Lecture 1 is correct, but not perfect):

by f'(x)=1 (1+x)+a=0 to get x=-1 a-1; And f''(x)=-1 (1+x) 2<0 is constant. So no matter if a takes any real number, x=-1 a-1 is the maximum point. However, to achieve a maximum at x=0, a=-1 is pushed out from x=-1 a-1=0.

f(x)=f(x)/a

x*lnx has a reputation for a

f'(x)=(1/a)[lnx+1]

f(x) has an extreme value at 1 e.

In (0,1 Stuffy e), f'(x) 0, f(x) monotonically decreasing;

In (1 e, ), f'(x) 0, f(x) monotonically increasing;

f(x) has a minimum at midroll 1 e.

When 0 1 e a, f(1 e) f(a) f(2a), the minimum value is f(a) = lna;

When 0 a 1 e 2a, the minimum value is f(1 e) = -1 (ae);

When 0 a, 2a, 1 e, the minimum value is f(2a)=2ln(2a);

Summary. Hello dear, I'm glad to answer 12It is known that the function +f(x)=ax-lnx-a x(1)If x>1, f(x)>0, the value range of the real number a is a>0(2) let x, x2 is) a>lnx1-x1 x2.

It is used in a variety of fields, including science, engineering, medicine, economics, and finance. Mathematicians also study pure mathematics, which is the substance of mathematics itself, without aiming at any practical application.

12.If x>1 is known to be the function +f(x)=ax-lnx-a(1)If x1, f(x)>0, the value range of the real number a is as: (2) Let x, x2 is.

Hello dear, I'm glad to answer 12It is known that the function +f(x)=ax-lnx-a x(1) If x>1, f(x)>0, the value range of the real number a is a>0(2) let x, x2 is the row stove) a>lnx1-x1 x2. It is used in a variety of fields, including science, engineering, medicine, economics, and Jinmeng Rongxue.

Mathematicians also study pure mathematics, which is the substance of mathematics itself, and does not aim at any practical application.

Extended Supplement: Mathematics is the result of the use of abstraction and logical reasoning, the study of counting, calculation, and the observation of the shape and movement of objects. Mathematics has become a part of education in many countries and regions.

f(x)=(1-x)/ax+lnx

a=1. f(x)=(1-x)/x+lnx

1/x+lnx-1

f'(x)=-1/x^2+1/x

(1/x-1/2)^2+1/4

Order f'=0, the solution is x=1

So when x [1 2,1), f'(x)<0,f(x) is a subtraction function when x(1,2], f'(x)>0, f(x) is the increment function, and when x=1 f'(x)=0 is the extreme point, and according to the above monotonicity, x=1 is the minimum point.

So, for the function f(x)=1 x+lnx-1, we can compare the values of f(1 2), f(2) points.

f(1/2)=2-ln2

f(2)=1 2+ln2-1=ln2-1 2f(1 2)-f(2)=5 2-2ln2>5 2-2>0, i.e. f(1 2)>f(2).

So the maximum value is 2-ln2 and the minimum value is 0

1) f’(x)=‘1/x - a/(x-1)^2 = x^2-(2+a)x+1] /x(x-1)^2]

Since the function f(x)=lnx+a (x-1) has an extreme value in (0,1 e), the function is derivable everywhere in this range.

So the derivative of the extreme point is zero.

So the molecule of the derivative x 2-(2+a)x+1 has a solution in the range (0,1 e).

4a+a 2 0 solution a Wu fiber-4, or a 0

In addition, it is necessary to ensure that the solution is within (0,1 e), since the axis of symmetry is 1+a2, if a -4 will lead to no solution.

So a 0 and f(1 e) 0 can be solved to a (e-1) 2 e and there is only one solution in (0,1) because the axis of symmetry is to the right of x=1.

2) At (0,1), f(x)<0, and only the noise seems to have an extreme point, which must be the maximum, and take a=(e-1) 2 e

The extreme point is x1=[2+a- (4a+a2)] 2 =1 e and the maximum value is f(1 e)=-e

At (1, positive infinity)f(x)>0, from the above analysis, it can be seen that the bending key also has a unique and minimum point in this interval and takes a=(e-1) 2 e

The extreme point is x2=[2+a+ (4a+a 2)] 2 =e and the minimum value is f(e)=2-1 e

So f(x2)-f(x1)>e+2-1 e

The word is too feasting to block and difficult to beat, and Hu Hu invites the landlord to see**. And sleepy.

(2) From a=1 to f(x)=ln(x+1)-x2-x from f(x)=-

52x+b, get ln(x+1)-x2+32

x-b=0 reams (x)=ln(x+1)-x2+32

x-b, then f(x)=-52

x+b has exactly two different real roots in the interval [0,2], which is equivalent to (x)=0, and there are exactly two different real roots in the interval [0,2]....(4 points) x) = 1x+1

2x+32−(4x+5)(x−1)2(x+1)

(5 points) when x [0,1], x) 0, then (x) increases monotonically on {0,1);

When x (1,2], x) 0, then (x) decreases monotonically on (1,2), according to the meaning of the title (0)=-b 0,(1)=ln(1+1)-1+32

b＞0，（2）=ln（1+2）-4+3-b≤0

solution, ln3-1 b ln2+12;

f(x)=f(x)/a

x*lnx/a

f'(x)=(1/a)[lnx+1]

f(x) has an extreme value at 1 e.

In (0,1 e), f'(x) 0, f(x) monotonically decreasing;

In (1 e, ), f'(x) 0, f(x) monotonically increasing;

f(x) has a minimum at 1 e.

When 0 1 e a, f(1 e) f(a) f(2a), the minimum value is f(a) = lna;

When 0 a 1 e 2a, the minimum value is f(1 e) = -1 (ae);

When 0 a, 2a, 1 e, the minimum value is f(2a)=2ln(2a);