Monotone interval of F (x) = in (square of 1 + X under x + radical)

Monotone interval of F (x) = in (square of 1 + X under x + radical)

Monotone increasing from negative infinity to positive infinity
For the function derivation, from (x + root sign (1 + x square)) > 0 can draw a conclusion!
Solution steps:
1. According to the meaning of the title, we first discuss the definition domain of y = f (x);
2. F '(x) is derived from F (x);
3. When f '(x) > 0, it is an increasing function and the corresponding interval is an increasing interval;
When f '(x)
Let f (x) = the square of x minus 4x + 3 under the root sign; (1) find the domain of definition of F (x); (2) find the monotone interval of F (x)
1）
X-4x + 3 ≥ 0
（x-1）（x-3）≥0
X ≤ 1 or X ≥ 3
That is, the domain of definition is (- ∞, 1] u [3, + ∞)
(2) X-4x + 3
=(X-2) m3-1
When x is less than or equal to 2, it decreases. Considering the domain, so
The minus interval is (- ∞, 1];
In the same way
The increasing range is
【3,+∞）.
(1) Domain of definition: x-4x + 3 > 0, x > 3 or X
F (x) = (3 times of TaNx under 1 + radical) / [1 + (TaNx) square], find monotone increasing interval
f(x) =(1+√3tanx)/(1+tan^2x).f(x)=1+√3tanx)/sec^2x.=(1+√3tanx)*cos^2x.=cos^2x+√3sinxcosx.=(1+cos2x)/2+√3/2*sin2x.=(1/2)cos2x+√3/2sin2x+1/2.=sin2xcos30+sin30cos2x+1/2.∴f(x)=sin(2x+π/6)+1/2.∵ 2k...
The equation of asymptote of hyperbola is 2x ± 3Y = 0 and passes through the point P (√ 6,2)
Let the hyperbolic equation be (x ^ 2) / 9 - (y ^ 2) / 4 = k, bring in the solution of point P, and get k = - 1 / 3. Then bring back K, and get the equation as (y ^ 2) / (4 / 3) - (x ^ 2) / 3 = 1
Upstairs, how to use the asymptote equation as the hyperbolic equation
Five squares of the same size, covering an area of 3125 square centimeters, calculate the side length of each board,
3125/5=625
625 radical = 25
So the side length of each board is 25 cm
Each area is 3125 △ 5 = 625 square cm = 25 × 25 square cm
So side length = 25 cm
25。
With 3125 △ 5 = 625, this is the area of a square, and the square root is the side length
Area of a square: 3125 / 5 = 625
Suppose: the side length of each square is x cm
x^2=625
x=25
It is known that x 1 and x 2 are two real roots of the quadratic equation (a-6) x & # 178; + 2 ax + a = 0
① Is there a real number a so that - X1 + x1x2 = 4 + x2 holds? If so, find out the value of A. if not, please explain the reason. ② find out the integer value of the real number a with (x1 + 1) (x2 + 1) as a negative integer
If the equation (a-6) x & # 178; + 2aX + a = 0 has two real roots, then the discriminant = 4A ^ 2-4a (a-6) = 24a > = 0, then a > = 0
For quadratic equation, there is a-6 ≠ 0, that is, a ≠ 6
1)x1+x2=-2a/(a-6), x1x2=a/(a-6)
From - X1 + x1x2 = 4 + X2, we get: x1x2 = 4 + X1 + x2
Substituting: A / (a-6) = 4-2a / (a-6)
a=4a-24-2a
a=24
So when a = 24, - X1 + x1x2 = 4 + x2 holds
2）(x1+1)(x2+1)=x1x2+x1+x2+1=-2a/(a-6)+a/(a-6)+1=6/(6-a)
To make the above expression a negative integer, we have 6-a
If the equation (a-6) x & # 178; + 2aX + a = 0 has two real roots, then the discriminant = 4A ^ 2-4a (a-6) = 24a > = 0, then a > = 0
For quadratic equation, there is a-6 ≠ 0, that is, a ≠ 6
1)x1+x2=-2a/(a-6), x1x2=a/(a-6)
From - X1 + x1x2 = 4 + X2, we get: x1x2 = 4 + X1 + x2
Substituting: A / (a-6) = 4-2a / (a-6)
A = 4a-24-2a = 0, a > = 0
For quadratic equation, there is a-6 ≠ 0, that is, a ≠ 6
1)x1+x2=-2a/(a-6), x1x2=a/(a-6)
From - X1 + x1x2 = 4 + X2, we get: x1x2 = 4 + X1 + x2
Substituting: A / (a-6) = 4-2a / (a-6)
a=4a-24-2a
a=24
So when a = 24, - X1 + x1x2 = 4 + x2 holds
2）(x1+1)(x2+1)=x1x2+x1+x2+1=-2a/(a-6)+a/(a-6)+1=6/(6-a)
To make the above expression a negative integer, we have 6-a
0.4x + 0.4x-24 + 56 = x the known solution is x = 160
First, merge the same category 0.8x + 32 = X
Then move the item (note to change the sign) 32 = x-0.8x
Then merge the similar terms (exchange the positions of the two sides of the equation without changing the sign)
0.2x＝32
Divide both sides by 0.2
x＝160
0.4x+0.4x-24+56=x
(0.4+0.4-1)x=24-56
-0.2x=-32
x=160
0.8x+32=x
0.2x=32
x=160
Second grade mathematics square root aspect question
1. Given that 20n is an integer under the root, how can the minimum positive integer n satisfy the condition be_____ .
20n=4*5*n
The root of 4 can be an integer
5 can't make the smallest positive integer n of 5 can be 5
Then the smallest positive integer that can make 20n open root integer is 5
N is 20
20=2²×5
Then the degree of all factors must be even
The number of 2 is even
As long as the number of 5 is even
Then just multiply by 5
So the minimum n is 5
The minimum positive integer is 5
Because 20n under the root sign is an integer, when 20n under the root sign is 1, n is not an integer. When 20n under the root sign is 2, n is not an integer. So when n is 5, 20n under the root sign is 10, so the minimum positive integer n is 5.
α. β is the real root of the equation x ^ 2-2ax + A + 6 = 0. Find the minimum value of (α - 1) ^ 2 + (β - 1) ^ 2
∵ the quadratic equation x2-2ax + A + 6 = 0 has two real roots;
∴△=4a²-4×（a+6）=4a²-4a-24≥0；
The solution is a ≤ - 2 or a ≥ 3;
∵ α, β are the two real roots of the quadratic equation x & # 178; - 2aX + A + 6 = 0 with respect to X;
∴α+β=2a,α•β=a+6；
（α-1）²+（β-1）²=α²+1-2α+β²-2β+1=α²+β²-2（β+α）+2
=（α+β）²-2αβ-2（α+β）+2
=4a²-2×（a+6）-2×2a+2
=4a²-2a-10
=4 (A-3 / 4) & 178; - 49 / 4;
∵ a ≤ - 2 or a ≥ 3;
(A-3 / 4) & # 178; ≥ (49 / 4) & # 178;;
4 (3 / 4 of A-4) & 178; - 49 / 4 ≥ 8;
Then the minimum value of (α - 1) &# 178; + (β - 1) &# 178; is 8
The equation has a real root, that is, 4A and 178; - 4a-24 ≥ 0
A ≥ 3 or a ≤ - 2
From the relationship between root and coefficient, α + β = 2A, α β = a + 6
　∴:(α-1)²+(β-1)²＝α²－2α＋1＋β²－2β＋1
　　　　　　　　＝（α²＋β²）－2（α＋β）＋2
= (α + β) & #178; -... Expansion
The equation has a real root, that is, 4A and 178; - 4a-24 ≥ 0
A ≥ 3 or a ≤ - 2
From the relationship between root and coefficient, α + β = 2A, α β = a + 6
　∴:(α-1)²+(β-1)²＝α²－2α＋1＋β²－2β＋1
　　　　　　　　＝（α²＋β²）－2（α＋β）＋2
　　　　　　　　＝（α＋β）²－2αβ－2（α＋β）＋2
　　　　　　　　＝（2a）²－2（a＋6）－2（2a）＋2
　　　　　　　　＝4a²－6a－10
Let t = 4A & # 178; - 6a-10
The vertex of the quadratic function is (3 / 4, - 49 / 4), which is not in the range of a, so the minimum value of the function can only be obtained at the endpoint. 　
When a = 3, t = 8; when a = - 2, t = 18, 8 < 18
The minimum value of T is 8. That is what we want. Put it away
The following equation? (2x-1) &# 178; - X & # 178; = 0?
（2x-1）²-x²＝0
(2x-1+x)(2x-1-x)=0
(3x-1)(x-1)=0
x1=1/3
x2=1
4x^2-4x+1-x^=0
3x^2-4x+1=0
(3x-1)(x-1)=0
X = 1 / 3 or 1
（2x-1）²-x²＝0
(2x-1+x)(2x-1-x)=0
x1=1/3
x2=1
(2x-1-x)(2x-1+x)=0
So (x-1) (3x-1) = 0
So x = 1 or x = 1 / 3