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F (x) defined on (0, π / 2), f '(x) is its derivative, and f (x) f (1) is greater than 2F (π / 6) sin1
F=f(x)/sinx
F'=[f'(x)sinx-f(x)cosx]/(sinx)^2>0
Add. 1 > pi / 6 to F, and replace it with F
It is known that f (x) is an increasing function defined on (0, + ∝), f (2) = 1, f (x1x2) = f (x1) + F (x2) is (0, + ∝) constant for X1 and X2, and the solution inequality f (x) is greater than or equal to 3 + F (X-2)
F (x) is defined on (0, + ∝)
F (X-2) is defined on (0, + ∝)
x-2>0
x>2
f(8)=f(2*2*2)=f(2)+f(2)+f(2)=1+1+1=3
f(x)≥3+f(x-2)
=f(8)+f(x-2)
=f[8(x-2)]
F (x) is an increasing function defined on (0, + ∝)
x≥8(x-2)
7x≤16
x≤16/7
So, 2
1. The tangent equation of curve y = x / (X-2) at point (1, - 1) is? 2. If the tangent inclination angle of curve y = 1 / X at a certain point (x1, Y1) is 135 degrees, X1 =? 3. Known curve C: y = x * x * X (1) Find the tangent equation at the point with abscissa 1 on curve C; (2) Is there any other common point between tangent and curve C in question (1)?
1、y'=[(x-2)-x]/(x-2)^2=-2/(x-2)^2 y'(1)=-2 y+1=-2(x-1) y=-2x+1
2、y'=-1/x^2 y'(x1)=-1/(x1)^2=tan135°=-1 x1=±1
3、y'=3x^2 y'(1)=3 y(1)=1 y-1=3(x-1) y=3x-2
There are some common points
To make an uncovered box with a rectangular bottom, its volume is 36cm ^ 3, the length ratio of the two adjacent sides of the bottom is 1:2, and what are the three sides (length, width and height) to minimize the surface area? To do this with derivatives,
If the length of the short side of the bottom is x and the height is y, then x * 2x * y = 36, so y = 18 / x ^ 2
S=x*2x+2*x*y+2*2x*y=2*x^2+36/x+72/x=2*x^2+108/x
Make the derivative of s to x DS / DX = 4 * x-108 / x ^ 2
If DS / DX = 0, there is
4*x-108/x^2=0
The solution is x = 3
So y = 18 / x ^ 2 = 2
Therefore, the length, width and height are 6, 3 and 2 respectively
Three topics with physics in Mathematics (the topic is in the chapter "derivatives and their applications") 1. If a train advances along a straight track and the train speed after braking V (T) = 27-0.9t (unit: M / s), how many m does the train stop after braking? 2. According to the law, x = 4T ² (unit: m) when moving in a straight line, assuming that the resistance of the medium is directly proportional to the speed, and the speed is 10m / s, the resistance is 2n, then what is the work done by the resistance from x = 0 to x = 4? 3. Hold a certain amount of gas in the cylindrical container with bottom area s. under isothermal conditions, due to the expansion of the gas, push a piston (area s) in the container from point a to point B to calculate the work done by the gas pressure during the movement If you know, say it quickly and write a few words,
1. S = (1 / 2) {[V (T) - 0] / T} * (square of T) = (1 / 2) (27-0.9t) t
Speechless, formula editing is too troublesome!
High school mathematics problems (with the application of derivatives) Known function f (x) = x ^ 3 + alnx (1) When a = - 2, find the monotone interval and extreme value of function f (x) (2) If G (x) = f (x) + 2 / X is a monotone function on [1, + ∞), find the value range of real number a
Very simply, I assume that you already know the formula expression of the root of univariate cubic polynomial
1) F (x) = x cube + alnx when a = - 2
F (x) = x cubic - 2lnx, then f '(x) = 3x square - 2 / x = (3x cubic - 2) / X
For univariate cubic polynomial 3x Cube - 2 = 0, according to the expression of the root of univariate cubic polynomial
Get 3x cube-2 = 3 [x-cubic root (2 / 3)] [x square + cubic root (2 / 3) + cubic root (4 / 9)]
It is easy to know that the latter term of the equation is obviously greater than 0
Therefore, we only need to calculate x [under x-cubic root (2 / 3)] and X ≠ 0
It is easy to judge that the extreme point x = under the cubic root sign (2 / 3), which is the minimum point, and the minimum value is f [under the cubic root sign (2 / 3)] = 2 / 3-2 / 3ln (2 / 3)
The monotone interval increases on (- ∞, 0), decreases on (0, under the cubic root sign (2 / 3)], and decreases on [under the cubic root sign (2 / 3), + ∞)
2) G (x) = f (x) + 2 / x = x cube + alnx + 2 / X is monotonic on [1, + ∞), then G '(x) = 3x square + A / X-2 / x square = [3x fourth power + AX-2] / x square, and its sign remains unchanged
Let u (x) = 3x fourth power + AX-2 obviously requires that its sign is also invariant, u '(x) = 12x cube + A in [1, + ∞)
If (A / 12) 12 + a > 0 under the triple root sign, then u '(x) > 0 3 + A-2 > 0 gets a > - 1
At this time - 1 (12 / 15) * four times under the root sign (120)
It is easy to find that a > is equal to 12 cubic meters
To sum up, a > - 1
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