High school math function problems, ask for expert solutions.

1.Proof:

Let's assume that f(x) = g(x) +h(x) exists, and let it be 1, then f(-x) = g(-x) +h(-x), and let it be the odd function property of 2: g(x) = -g(-x).

Even function properties: h(x)=h(-x).

Then take 1 + 2 and 1 - 2 respectively to get:

f(x)+f(-x)=2h(x)

f(x)-f(-x)=2g(x)

From this we conclude that for any f(x), we can construct the following two functions: g(x)=[f(x)-f(-x)] 2 is an odd function h(x)=[f(x)+f(-x)] 2 is an even function and g(x)+h(x)=f(x).

Certification. 2.Let f(x) and g(x) be even functions, so that f(x) = f(x) + g(x).

then f(-x)=f(-x)+g(-x)=f(x)+g(x)=f(x), so the sum of two even functions is an even function;

In the same way, f(x) and g(x) can be odd functions, so that f(x)=f(x)+g(x) then f(-x)=f(-x)+g(-x)=-f(x)-g(x)=-f(x), so the sum of two odd functions is an odd function;

3.As with 2, let f(x) and g(x) be used to prove it according to the properties of the function.

1) Here you need to have f(x) to define the domain as symmetric, there is f(x) = f(x) + g(x).

f(x)=[f(x)+f(-x)] 2 is an even function.

g(x)=[f(x)-f(-x)] 2 is an odd function.

2) Let f(x), g(x) be even.

then f(x) = f(x) g(x).

f(-x) = f(-x)g(-x)=f(x)g(x)=f(x) so f(x) is an even function.

The same is true of others.

3) the same as 2) method.

Because the curve is getting slower and slower to remove the empty base, the slope of its tangent is getting smaller and smaller, so the slope of 2 is greater than 3, according to the image you draw, f3 f2 is larger than the slope of 2, and the brigade chooses c

2 f (x) 3x deficit 2 6a

According to f(x), the monotonicity of f(x) can be judged to be an increase or decrease, and then the extreme values are discussed.

Calculate the range of a.

Let -1f(-x)=4 x 2 x+1=-f(x)f(x)=-2 x-1

Hongshi 2).f(x)=1/2^x+1

2 x increases as x increases.

So f(x) is monotonically reduced on the hole (0,1).

2^|x|>=1

1+a>=1.

a>=0

When a=0, x=0 has a solution.

a>0 two solutions.

a<0 There is no solution.

Drawing is the most straightforward way to see how many intersections there are.

1. High family solution: about origin symmetry: He Wu let a (Qi beat cheat x,y) is the function at any point on (-1,0), then a'(-y,-x) is the point on f(x)=2 x 4 x+1, then there is:

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

Write y as a function of x, just summarize it. [Because there may be a problem with your question (I don't know if there are parentheses in the denominator), the answer is not easy to write, so write the idea!] 】

2) Derivatives can be found, or variables can be separated.

2. Tree combination:

2^|x|-1>=0 and symmetrical with respect to the y-axis.

Easy to know: when a<0, there is no solution;

a=0, the only solution;

a>0, two solutions.

With respect to origin symmetry, it is the odd silver function, there is f(x)=-f(x)Let x belong to (0,1), then -x belongs to (-1,0), and bring in the hood ridge f(x)=(1 2) x) (1 4) x-1).x belongs to (-1,0), and f(x)=2 x the number of objects infiltrates 4 x+1

2), let 2 x=a, then f(x)=a a 2+1=1 (a+1 a). Let g(x)=a+1 shuyu (1,2)diu g(x)qiu dao,de a zai (1,2),f(x) zai (0,1)dandiao dizeng.

2,2^jueduizhi x=a+1 .Put f(x)=2 juediuzhi x de tuxiang hua chuAvailable to dang a xiaoyudengyu-1shi you meiyougeng ,dang a dayu -1 shi you 1ge geng

Odd function, a is equal to zero one solution, less than zero no solution, greater than zero two solutions.

2) The original formula is f(x)=(1 2) x+(1 3) x m, and it is easy to know that f(x)=(1 2) x+(1 3) x is monotonically decreasing, so the minimum value of f(x) in the interval (- code file 1) is f(1)=1 2+1 晌碧3=5 6, so the value range of m is (- 5 6].

It can be very vividly obtained from the image.

The function is symmetrical with respect to the center of (-1,1) points.

Increase interval (-1, +infinite).

(1) Solution: according to the topic.

f（x）=x/（1+x）=（1+x-1）/（1+x）=1-1/（1+x）

Pull f(x)=-1 x to the left and 1 unit upwards to get f(x)=x (1+x).

Only 1+x≠0

x≠-1 sets x1 x2 (x1, x2≠0), then.

f（x2）-f（x1）

x2/（1+x2）-x1/（1+x1）

x2（1+x1）-x1（1+x2）]/[（1+x2）（1+x1）]

x2-x1) [(1+x2)(1+x1)] when x1 x2 -1.

x2-x1＞0，1+x2＜0，1+x1＜0∴f（x2）-f（x1）＞0

f（x2）＞f（x1）

At this point, the function increases monotonically.

When -1 x1 x2.

x2-x1＞0，1+x2＞0，1+x1＞0∴f（x2）-f（x1）＞0

f（x2）＞f（x1）

At this point, the function increases monotonically.

f(x)=1-1 (x+1), you can draw this, right?

How do you pan do you know no?

The rest of the problem is easy to solve, we know that 1 x monotonic is a monotonic decreasing function, and the monotonic interval is (negative infinity, 0) and (0, positive infinity), so f(x) is a monotonically increasing function, and the monotonic interval is (negative infinity, -1) and (-1, positive infinity).

If you have any questions, please feel free to ask.

Good luck with your studies.

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

The image of f(x)=1 x is drawn first, then the whole is shifted to the left by 1 unit, and the whole is flipped up and down along the x-axis, and finally the whole is moved up by 1 unit.

The monotonic reduction of f(x)=1 x is (negative infinity, 0) and (0, positive infinity). So the monotonic increase of this function is (-1, positive infinity) and (negative infinity, -1).

fx is equal to (1+x-1) divided by (1 plus x) equals 1 minus 1 divided by (1 plus x) compared to standard y is equal to 1 divided by x.

According to the present y, it is the original y plus 1

The current x is the original x plus 1 to get it.

As shown in the figure, this function is obtained by deforming f(x)=1-1 (1+x), that is, by translating 1 unit to the left and then 1 unit upwards by f(x)=1-1 x

Since its range is [0,+infinity), we can get all the numbers that x 2+2ax-a can get to 0 (instead of getting x 2+2ax-a 0 constant).

It should be 0, and only when 0 can the quadratic function with the opening upwards take all the positive numbers.

The function range is [0, + then 2 (x 2+2ax-a) is greater than or equal to 1, that is, x 2+2ax-a is greater than or equal to 0, so > 0, the solution is obtained.

Is your question f(x)=1 (1-x), g(x)=(ax+1) x??It's the other way around.