Help with two math problems Thank you

Question 1: The answer is (-1,0), i.e. the open range is -1 to 0

The process is as follows: since p and q are symmetrical with respect to the origin, the analytic formula for g(x) is to replace x in f(x) with -x and y with -y, i.e., -y=loga(-x+1), y=-loga(-x+1), thus.

g(x) =-loga(-x+1), so that 2f(x)+g(x) loga{(x+1) 2}-loga(-x+1)=loga>=0, due to 00].

In order for 0<(x+1) 2 (-x+1)<=1 to hold, its denominator -x+1>0 is given as equation (1) and (x+1) 2<=(-x+1) is given as equation (2) and x+1>0 is used as equation (3), and (1)(2)(3) is joined together to give that the range of x is (-1,0).

The answer to the second question is (-4,1), and the process is:

Let x=2, y=1, then there is f(2 1)=f(2)-f(1), f(1) 0 can be obtained, and also let x=4, y=2, f(4)=2, from f(x y)=f(x)-f(y), let z=1 y, then f(xz)=f(x)-f(1 z)=f(x)-f(x)=f(x)+f(z), So the multiplication is also defined, f(x+3)-f(1 x) f(x+3)-f(1)+f(x)=f(x+3)+f(x)=

f<2, and 2 f(4), so f solves the x range to be (-4,1).

Quite tired! Good luck!

Dude: What a complicated question.

1 point is not added to others.

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I don't want to earn your 2 points.

What I said upstairs is so right, there must be someone who will give a reward for such a complex question.

I look too dizzy, let's answer someone else!

Solution: The speed of the boat when going downstream is v=a+b The speed of the boat when going against the current is v=a-b according to the time = distance speed.

Therefore t=s (a-b)-s (a+b) =2bs (a -b )1 (1-x)+1 (1+x)=(1+x+1-x) (1-x)=2 (1-x).

So. 1 (1-x)+1 (1+x)+2 (1+x squared)+4 (1+x to the fourth power).

2 (1-x) + 2 (1+x )+4 (1+x to the fourth power) = 4 (1-x to the fourth power) + 4 (1+x to the fourth power) = 8 (1-x to the 8th power).

In the same way, you can find the second question and the second question.

The result is 16 (1-x to the 16th power).

1.The time from the upstream port to the downstream port tshun = s (a+b).

The time from the downstream port to the upstream port t inverse=s (a-b).

So the time difference between the two ports t=t inverse-t cis=s (a-b)-s (a+b).

2.Calculate 1 (1-x) + 1 (1+x) first, and get 2 (1-x squared) by passing the score

So we can get a formula 1 (1-y) + 1 (1+y) = 2 (1-y squared) here to distinguish us from x, we use the unknown quantity y

So when y = x squared, we take in the formula and it becomes 1 (1-x squared) + 1 (1+x squared) = 2 (1-x to the 4th power).

In the same way, when y=x to the power of 4 ...

We can use the conclusions obtained above to easily calculate these two questions.

First use the original formula and then propose the common multiple, then use the conclusion of y=x when the square and the third term are combined, and so on, knowing that I can't use a formulator, I'm tired to death.

The problem of time difference 842105620 already good.

The next question is actually quite simple:

1.8 (1-x to the 8th power).

2.16 (1-x to the 16th power).

1. Downwater time = s (a+b).

Time against the water = s (a-b).

So round-trip time = s (a+b)+s (a-b (1-x)+1 (1+x).

2/(1-x^2)

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

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

8/(1-x^8)

1/(1-x)+1/(1+x)+2/(1+x^2)+4/(1+x^4)+8/(1+x^8)

2/(1-x^2)+2/(1+x^2)+4/(1+x^4)+8/(1+x^8)

4/(1-x^4)+4/(1+x^4)+8/(1+x^8)=8/(1-x^8)+8/(1+x^8)

16/(1-x^16)

Find the speed in the current When going downstream—v=a+b," when going against the current—v=a-b

Therefore t=s (a-b)-s (a+b).

1. By Vedic theorem, x1+x2=--a, x1x2=a--2, so.

x1--x2) 2=(x1+x2) 2--4x1x2=29, so a=--3 or 7. When a=--3, there is a minimum value of --29 4, and when a=7, it is also --29 4. , 2, by Vedic theorem, x1+x2=3, x1x2=m, s=x1 2+x2 2=(x1+x2) 2--2x1x2=9--2m, by the discriminant formula m<9 4

s=7, by substituting x1 2=3x1--1, so.

x1 3 = 3x1 2--x1, substituting 21

1.Let the two roots be x1 and x2then (x1-x2) = 29i.e. x 1 + x 2-2 * x 1 * x2 = 29. i.e. (x1+x2) -4x1*x2=29

and x1+x2=-a, x1*x2=a-2

Substituting the above equation yields a=-3 or a=7. When a=-3, the minimum value of the function is -29 4When a=7, the minimum value of the function is also -29 4The minimum value is -29 4

Question 1: Because x ( 2,2) and xe 丨x丨 is an odd function in this interval.

e 丨x丨 is even, then.

Original 2 e xdx,x (0,2).

2e x+c, the generation interval can be calculated.

Question 2: Ling x rcos, y rsin

The original formula re (r 2)drd can be calculated only if the interval of the era into r.

Hope it works for you.

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