(1) Y = 3x + 2, X ∈ R (2) y = x & # 178; - 1 (x ≥ root 2) please find out the range of its function respectively,

(1) Y = 3x + 2, X ∈ R (2) y = x & # 178; - 1 (x ≥ root 2) please find out the range of its function respectively,

y=3x+2,x∈R
Because x ∈ R
So 3x ∈ R
So 3x + 2 ∈ R
y∈R
That is, the range ∈ R
y=x^2-1（x≥√2）
So x ^ 2 ≥ 2
So x ^ 2-1 ≥ 1
y≥1
That is, the value range is greater than or equal to 1

Find the range of F (x) = 12 / x + 3x

f（x）=12/x+3x,
|f(x）|=12/|x|+3|x|>=12,
The range of F (x) is [12, + ∞) ∪ (∞, - 12]

For function y = 2x + 1 / 1-3x definition field and value field! Request solution idea! Please experts help, thank you very much!

1-3x≠0 ,x≠1/3
x≠-2/3

The range of function y = 2x ^ 2 + 3x-6 (x belongs to d) is [1,4], and the domain D is obtained

That is to solve inequality group 1

Finding the monotonicity of the definition range of the function y = 1 / √ (x ^ 2 + 3x-4)

X ^ 2 + 3x-4 = (x-4) (x + 1) domain x ＞ 4 or X ＜ - 1
= (x-3 / 2) & # - 25 / 4 range (0, + ∞)
X ^ 2 + 3x-4 decreases monotonically when x ＜ - 1, and Y rises monotonically
When x > 3, it rises monotonically and Y decreases monotonically

Let the sum of the first n terms of the sequence {an} be Sn, and the points (n, Sn / N) (n belongs to n positive) are all on the image of function y = 3x-2
Let BN = 3 / ana (n + 1), tn be the sum of the first n terms of the sequence {BN},
Find TN

∵ points (n, Sn / N) (n belongs to n positive) are all on the graph of function y = 3x-2
That is, Sn = 3N ^ 2 - 2n
Then: s (n-1) = 3 (n-1) ^ 2 - 2 (n-1) = 3N ^ 2 - 8N + 5
By subtracting the two formulas, Sn - S (n - 1) = 6N - 5
That is: an = 6n-5
Then a (n + 1) = 6 (n + 1) - 5 = 6N + 1
bn=3/AnA(n+1) =3/(6n-5)(6n+1)
=3*(1/6)*[1/(6n-5) - 1/(6n+1)]
=(1/2)*[1/(6n-5) - 1/(6n+1)]
Then B1 = (1 / 2) * (1 / 1 - 1 / 7)
b2=(1/2)*(1/7 - 1/13)
…
…
bn=(1/2)*[1/(6n-5)-1/(6n+1)]
∴Tn=b1+b2+b3+.+bn
=(1/2)*{(1/1 - 1/7)+(1/7 - 1/13)+.+[1/(6n-5) - 1/(6n+1)]}
=(1/2)*[1-1/(6n+1)]
=3n/(6n+1)

Given the function f (x) = (2x + 3) / 3x, the first term of the sequence {an} A1 = 1, a (n + 1) = f (1 / an), the sum of the first n terms is SN
If the sum of the first n terms of the sequence {1 / 3Sn} is TN, for N, it belongs to n * and TN is less than f (m), then the value range of the real number m is obtained

A (n + 1) = f (1 / an) = (2 / an + 3) / (3 / an) = 2 / 3 + an, so {an} is an arithmetic sequence with 1 as the first term and 2 / 3 as the tolerance. An = (2n + 1) / 3Sn = (N2 + 2n) / 31 / 3Sn = 1 / (N2 + 2n) = 1 / N (n + 2) = 1 / 2 * (1 / n-1 / (n + 2))
Tn=1/2*(1-1/(n+2))

Given that f (x) = 4 + 1 / x2 under the root sign, the sum of the first n terms of the sequence {an} is Sn, the point PN (an, - 1 / an + 1) (n belongs to n *) is on the curve y = f (x), and A1 = 1, an > 0
(3) To prove 4N + 1-1 under Sn greater than 1 / 2 radical

1 / 2 (radical (4N + 1) - 1) is regarded as the sum of the first several terms of the sequence
TN = 1 / 2 (radical (4N + 1) - 1)
TN-1 = 1 / 2 (radical (4n-3) - 1)
BN = 1 / 2 (radical (4N + 1) - radical (4n-3))
Because 4 / (4n-3) + 4n-3 + 4 > 4N + 1
So it is the same as open radical, that is, 2 radical (1 / 4n-3) + radical (4n-3) > radical (4N + 1)
That is root (1 / 4n-3) > 1 / 2 (root (4N + 1) - root (4n-3))
That is, an > BN, that is, Sn > TN
So Sn > 1 / 2 (radical (4N + 1) - 1)

Given that f (x) = root 4 + 1 / X2, the sum of the first n terms of the sequence {an} is Sn, the point PN (an, 1 / an + 1) (n belongs to n *) is on the curve y = f (x), and A1 = 1, an > 0
Given that f (x) = root 4 + 1 / X2, the sum of the first n terms of the sequence {an} is Sn, the point PN (an, 1 / an + 1) (n belongs to n *) is on the curve y = f (x), and A1 = 1, an > 0

1 / an + 1 = √ (4 + 1 / an ^ 2) A2 = √ 5 / 51 / (an + 1) ^ 2 = 4 + 1 / an ^ 21 / (an + 1) ^ 2-1 / an ^ 2 = 4, so TN = 1 / (an + 1) ^ 2 is the arithmetic sequence T1 = 5Tn = 5 + (n-1) * 4 = 1 / (an + 1) ^ 2 = 4N + 1, so an + 1 = √ [1 / (4N + 1)] an = √ [1 / (4n-3)] (n > = 1)

It is known that in the positive term sequence {an}, A1 = 1, and the point P (an, Sn) (n ∈ n +) is on the function y = (X & sup2; + x) / 2 image
① Find the general term formula of sequence {an}
② Let BN = 1 / an and Sn denote the sum of the first n terms of the sequence {BN} +Is s (n-1) = sn-1 * g (n) constant for all natural numbers n not less than 2?
If it exists, write the analytic expression of G (n) and prove it; if not, explain the reason

Point P (an, Sn) (n ∈ n +) in the function y = (x ^ 2 + x) / 2 image, we can see Sn = (an ^ 2 + an) / 2, so 2Sn = an ^ 2 + an ① 2S (n-1) = a (n-1) ^ 2 + a (n-1) ② subtracting the above two expressions to obtain an + a (n-1) = an ^ 2-A (n-1) ^ 2, dissolving an-a (n-1) = 1, that is, an is an arithmetic sequence with tolerance of 1. And A1 = 1, we can obtain an = NbN = 1 / n