Let the function f x ln x 1 ae x a, a belongs to R

1) If a=1, f(x)=ln(x+1)-e (-x)-1, x>0, let x1 be less than x2, and bring in the monotonicity.

This is the method of definition.

You can also look directly at the monotonicity of the function.

ln(x+1) is an increasing function e (-x) is a decreasing function so -e (-x) is an increasing function an increasing function or an increasing function.

f(x)↑。

2) ln(x+1)+ae(-x)-a>=0(x>=0), x=0;

1-e (-x) >0 at x>0, a<=ln(x+1) [1-e (-x)], denoted as g(x), g'(x)= [1-e (-x)] 2, let h(x)=e x-[1+(x+1)ln(x+1),x>0, then.

h'(x)=e^x-[ln(x+1)+1],h''(x)=e^x-1/(x+1)>0,h'(x)↑，h'(x)>h'(0)=0,h(x)↑，h(x)>h(0)=0,g'(x)>0,g(x),g(x)>g(0)=0,In summary,a<=0.

Proving monotonous, derivatives can be sought.

After that, there is the second question, which is to treat a as an unknown variable and x as an argument, and look at the inequality of f(x)>=0.

1）f(x)=ln)x+1)-e^(-x)-1,x>0,f'(x)=1/(x+1)+e^(-x)>0,f(x)↑

2) ln(x+1)+ae(-x)-a>=0(x>=0), x=0;

x>01-e^(-x)>0,a0, h'(x)↑h'(x)>h'(0)=0,h(x) h(x)>h(0)=0,g'(x)>0,g(x)↑

g(x)>g(0)=0, synthesis a,6, let the function f(x)=ln(x+1)+ae (-x)-a, a belongs to r (1), when a=1, it is proved that f(x) is in.

0, positive infinity) is the increasing function (2), if x belongs to [0, positive infinity), f(x) is greater than equal to 0, find the range of the hail value of a.

Summary. Hello, I'm glad this question was left to me. When a=1, find the tangent equation for the curve y=f(x) at the point (2,f(2)).

The ordinate of the point (2, f(2)) is f(2)=1 2+ln2-1=ln2-1 2

Derivative f'(2)=-1/2^2+1/2=1/4

The tangent equation y-f(2)=f'(2)*(x-2)

i.e. y=1 4*x+ln2-1

Let the function f(x)=ae x-lnx-1, when a is greater than or equal to one part of e, dear, hello, glad that this slippery luggage problem is up to me. In order to better help you, please ask the lead where you saw this question, you can take a picture to show me the question, thank you!

Hello, I'm glad this question was left to me. When a=1, find the tangent equation of the curve y=f(big cherry x) at the point (2,f(2)) The ordinate of the point (2,f(2)) is f(2)=1 2+ln2-1=ln2-1 2 derivative f'(2)=-1 2 2+1 2=1 4 tangent equation rolling bush y-f(2)=f'(2)*(x-2)i.e. y=1 4*x+ln2-1I hope mine can help you, I wish you a happy life!

Summary. If a=1, the tangent equation for finding the image of the function f(x) at (1,f(1)) is y=(e-1)x-1

The known function f(x) = ae'-lnx-2(aer).+1) If a=1, find the tangent of the image of the function f(x) at (1,f(1)).

If a=1, the tangent equation for finding the image of the function f(x) at (1,f(1)) is y=(e-1)x-1

Using the derivative method, the slope of the tangent is found.

What about the second question? 2.[Elective 4-4:.]

Coordinate System and Parametric Equation] (10 points) In the Cartesian coordinate collapse system XOY, the coordinate origin O is the pole, and the positive semi-axis of the x-axis is the polar axis, and the polar coordinate equation of the curve c is p, and the polar group of the curve c is p=4cos 0+2sin 0(1) Combine curve c and curve c. The polar coordinate equation is converted into a Cartesian coordinate equation; (2) If curve c and curve c.

Intersect at two points m, n, disadvantage and defeatmn|.

The second question is the constructor, finding the derivative, and determining the minimum value.

There are two derivatives to be taken, assuming that the zero point is x0, and finally using the fundamental inequality to reach a conclusion.

2.[Elective 4-4: Coordinate System and Parametric Equation] (10 points) In the Cartesian coordinate collapse system XOY, the polar coordinate system is established with the coordinate origin o as the pole and the positive semi-axis of the x-axis as the polar axis, the polar coordinate equation of the curve c is p, and the polar group of the curve c is p=4cos 0+2sin 0

(1) Combine curve c and curve c. The polar coordinate equation is converted into a Cartesian coordinate equation; (2) If curve c and curve c. Intersect at two points m, n, disadvantage and defeatmn|.

Help me get to the point.

In the first question, follow the steps of converting polar coordinates into Cartesian coordinate equations.

2√2ρ²=8x²+y²=8

The second one is the same, first squared.

The second multiplies on both sides

The polar coordinate equation is prepared into a Cartesian coordinate equation steps: the polar coordinate equation is arranged into the form of cos and sin; Roll to convert cos into x and sin into y; Replace "" with (x y). x=ρcosθ，y=ρsinθ，x²+y²=ρ

f'Row slag rent (x) = 1-ae x

When the file is 0, f'Liang Hunger (x) >0

f(x) monotonically increases, f(0)=1-a>0, which does not meet the topic.

Therefore a>0, let f'(x)=0

1=ae^x

e^x=1/a

x = ln(1 a) = -lna when x

The value of the real number a is found by knowing that the functions f(x)=ae x, g(x)=lnx-lna, where a is a constant, and the tangents of the functions y=f(x) and y=g(x) are parallel to each other at the intersection of the two axes.

Solution: f(x)=ae x, f(0)=a, intersection with y-axis (0,a), f(x)=ae x, f(0)=a;

g(x)=lnx-lna, g(a)=lna-lna=0, intersection with x-axis (a,0), g(x)=1 x, g(a)=1 a;

Since the tangent of the image of y=f(x) and y=g(x) is parallel to each other at the intersection point of the two coordinate axes, there is a=1 a, that is, a =1 and a=1

f′(x)=aex，g′(x)=1

The image of x y=f(x) intersects the coordinate axis at the point (0,a); The image of y=g(x) intersects the coordinate axis at the point (a,0), f (0)=g (a) a=1a

a＞0，∴a=1

g（x）=lnx．

1）x∈(0,+∞

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

When f(x).'=0 x=a

x→0+f(x)→

f(a)=lna

f(e)=a/e

When a (-0), there is no minimum value.

When a (0,e], the minimum value: f(a) = LNA when a (e,+ minimum value: f(e) = a e2) i.e. g(x).'=0

g(x)'=ln(x)-1)e^x + e^x)/x +1g(x)'‘ln(x)-1) e^x - e^x)/(x^2) +2 (e^x)/x

g(x)''0 x=1

x Flush 0+ g(x).'Cong Juqin +

g(1)'=1

x→+∞g(x)'→

f(x)'Evergrande is 0, so it is not zero.

So it doesn't exist.

It is easy to know that f(x)g(x), the value range is that x>0, and the derivative function of (1) f(x) is: f'(x)=-a x 2+(1 x) let f'(x)>=0, and x>=a

f'(x)<0, then the minimum value of f(x) is: f(e)=a e;

2）g’(x)=(1/x)e^x+(lnx-1)e^x+1

f(x)'=-a x 2+1 x discuss the range of values of a to determine the monotonic interval.

Perpendicular to the y-axis, the slope of the tangent is 0, g(x).'=(1/x-1)e^x+（lnx-1）e^x+1=0

Find out whether it has a solution in the interval (0,e).