Monotone interval of function y = root sign (square of x-2x-3)

Monotone interval of function y = root sign (square of x-2x-3)

Y = root sign (x ^ 2-2x-3) = root sign [(x-1) ^ 2-4]
Definition field: (x-1) ^ 2 ≥ 4, X ≤ - 1 or X ≥ 3, axis of symmetry x = 1
In (- ∞, - 1] interval, (x-1) ^ 2 monotonically decreases, y = root sign [(x-1) ^ 2-4] monotonically decreases;
In [3, + ∞) interval, (x-1) ^ 2 increases monotonically, and y = root sign [(x-1) ^ 2-4] increases monotonically

What is the minimum value of function y equal to the square of X under the root plus 2x plus 2 plus the trivial minus 4x plus eight of X under the root?

2 √ 5. In fact, as long as the binary primary equations on both sides are solved, the binary primary equations on both sides are equal, solve x = 1, and bring x = 1 into the function is the minimum value

Y is equal to the square X of COS root x plus Xe, find y. how does this problem work? wait anxiously Seek dy

Your question is not clear, especially the mathematical symbols, which is easy to cause ambiguity: y = cos √ x + Xe ^ (2x) derivation, y '= - sin √ X / (2 √ x) + e ^ (2x) + 2xe ^ (2x), so dy = y' DX = [- sin √ X / (2 √ x) + e ^ (2x) + 2xe ^ (2x)] DX solution to the problem: it's

Y = ln [arctan (1-x)], which is dy=

y'=1/arctan(1-x)*1/[1+(1-x) ²]* (1-x)'
=-1/[(x ²- 2x+2)arctan(1-x)]
therefore
dy=-1/[(x ²- 2x+2)arctan(1-x)]dx

It is known that [x] represents the largest integer not exceeding the real number x, G (x) = [x] is the rounding function, and x0 is the function f (x) = LNX − 2 Zero of X, then G (x0) is equal to () A. 1 B. 2 C. 3 D. 4

∵f(2)＝ln2−1＜0，f(3)＝ln3−2
3＞0，
So x0 ∈ (2,3),
∴g（x0）=[x0]=2．
So choose B

For all real numbers x, let [x] be the largest integer not greater than x, then the function f (x) = [x] is called a Gaussian function or rounding function, and calculate f (- 0.3) + F (1) + F (1.3) = ___; If an = f (n / 3), n ∈ n +, Sn is the sum of the first n terms of the sequence {an}, then s3n = ___

f(-0.3)+f(1)+f(1.3)=-1+1+1=1,
S3n=0+0+1+1+1+2+2+2+...+(n-1)+(n-1)+n=3(1+2+...+(n-1))+n
=(3n-1)n/2