If there is an intersection between the original function and the inverse function, is the intersection on the straight line y = x? Why? He also gave a counter example, but I forgot. Now I'm in college. I'm confused when I encounter such problems,

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I didn't understand what you wrote
Which root?
Generally, the first step of such a problem is to find the definition domain and range, and then solve it
What I solved is (X-9) - 1 open root. It seems that there is a solution. I don't know what your problem is
The definition domain is {x | x > 10}
The range is r
After dinner, I'll come back
The domain is x > 9
The range is r
By the way, it's x plus 6 times the root, X-9, and finally subtract 1, right
I can't understand the problem
Ah, ah, I hate to use computers to talk math problems

Exercise 1.3 question 6 of group A When x ≥ 0, f (x) = x (1 + x) draws the function image and obtains the function analytic formula

A function is odd on R
There are: F (- x) = - f (x),
When x ≥ 0, - x

24 pages 2 3

Two point four
3.cosB=(a-bcosα)/[√(a^2+b^2-2abcosα)]
cosA=(b-acosα)/[√(a^2+b^2-2abcosα)]
5. The river width is Hsin (α - β) / sin α * sin β

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Case 1: because a belongs to R, AUB = {1,3,4} a intersection B = {1} case 2: when a = 4 AUB = 1 34 a ∩ B = 4: when a = 3 AUB = 1 34 a ∩ B = empty set case 4: a ≠ 1 34 a ∩ B = empty set AUB = 1 34 a

2 page 24 group A Question 2 （1）f（x）=x-1,g（x）=x/x2-1 （2）f（x）=x2,g（x）=（√x）4 （3）f（x）=x2,g（x）=3√x6

Is f (x) and G (x) the same function
(1) No, because g (x) = x2 / X-1, the domain of definition is x not = 0, while the domain of F (x) is all real numbers R
(2) No, because the domain of G (x) is x > = 0, and the domain of F (x) is r
(3) Yes

Emergency treatment within 24 hours It is known that the sum of the first three terms of an is 6, and the sum of the first eight terms is - 4 1） The general formula of an 2) Let BN = (4-an) QN-1 (Q ≠ 0, n ∈ n *), find the first n terms and Sn of the sequence BN

1) ∵ an is the arithmetic sequence
∴an=a1+(n-1)d
Sn=na1+[n(n-1)/2]d
∴S3=3a1+3d=6………… (1)
S8=8a1+28d=-4……… II.
→a1=3 ,d=-1
∴an=4-n
2）bn=(4-4+n)qn-1
=qn^2-1
∵q≠0,n∈N*
∴Sn=qn(n+1)(2n+1)/6-n

1. Given the vector a = (cosx, SiNx), B = (2cos (x / 2), - 2Sin (x / 2)), and X ∈ (- π / 9,2 π / 9] (1) Find the value range of A.B and | A-B | (2) Find the minimum value of the function f (x) = A.B - | A-B | 2. If a = (1,3), B = (2, λ), let the angle between a and B be θ, and find the value range of λ so that θ is an acute angle 3. The three sides a, B and C of the triangle ABC are integers with a perimeter of 20 and an area of 10 √ 3. The three inner angles a, B and C form an arithmetic sequence to find the length of the three sides of the triangle

1. 1. 1. A.B = cosx * 2cosx / 2-sinx * 2sinx / 2 = 2cos3x / 2-sinx * 2sinx / 2 = 2cos3x / 2 | A-B | = a | - 2A B + B | B | a | a | - 2A B + | B | B | B | B | [1,3] two sides open square ∈ [1,3] the line is square on both sides, ② f (x) = A.B - | A-B = 2cos3x = 2cos3x / 2 | (5-4cos 3x / 2) let t = cos 3x / 2, then t = cos 3x f (x) = 2T - √ (5-4t)

Help solve the math problem of the next senior one In triangle ABC, sinA:sinB : sinc = m: (M + 1): 2m to find the value range of M. (detailed process)

Let a, B and C be the three sides of a triangle. From the sine theorem, a: Sina = B: SINB = C: sinc
sinA:sinB :sinC=a:b:c
So: A: B: C = m: (M + 1): 2m
Because the sum of any two sides is greater than the third
So (a + C): b > 1
That is + 1 / M > 2m
In the same way, we can get: M > - 1 / 2
So: M > 1 / 2

1. If 2lg (x-2y) = lgx + lgY, then the value of X / y is () A1 B4 C1 or 4 D1 / 4 or 4 2. Let x ∈ R, if a < LG (︱ x-3 ︱ x + 7 ︱) holds, then () A a ＞ equals 1 B a ＞ 1 C 0 ＜ a ＜ equal to 1 d a ＜ 1 3. Let f (x) = f (1 / x) lgx + 1, then the value of F (10) is () A 1 B -1 C 10 D 1/10 4. If the function f (x) = log2a (x + 1) defined in the interval (- 1,0) satisfies f (x) > 0, then the value range of a is () A (0,1 / 2) B (0,1 / 2] C (1 / 2, positive infinity) d (0, positive infinity) 5. The function f (x) = 1 / 2 (AX + A-X) (x - x is the power) (a > 0 and a is not equal to 1) image passing through point (2,41 / 9) (1) It is proved that f (x) is an increasing function on [0, positive infinity]

1. Solving the logarithmic equation: (x-2y) ^ 2 = XY
x^2-5xy+y^2=0
X = 4Y or x = y
So, X / y = 4 or 1
2. The two sides of the inequality are represented by the logarithm of the same base
LG (10 ^ a) 1, then the image of F (x) is negative on (- 1,0), which does not meet the requirements of the title,
Only if 0

If there is an intersection between the original function and the inverse function, is the intersection on the straight line y = x? Why? He also gave a counter example, but I forgot. Now I'm in college. I'm confused when I encounter such problems,