The function f (x) = ax ^ 3 + BX ^ 2 + CX + 3 defined on R satisfies the following conditions 1. F (x) is a decreasing function on (0,1) and an increasing function on (1, + OO) 2. F '(x) is even function 3. The tangent of F (x) at x = 0 is perpendicular to the line y = x + 2 Finding the analytic expression of function y = f (x) Let g (x) = 4lnx-m, if x belongs to [1, e], Let g (x)

(1)
Because f (x) is an even function, f (- x) = f (x)
We obtain 4bx = 0, that is, B = 0
Because the tangent of F (x) at x = 0 is perpendicular to the line y = x + 2
So f '(0) * 1 = - 1
Get C = - 1
Because f (x) is a decreasing function on (0,1) and an increasing function on (1, + OO)
When x = 1, f '(x) = 3ax ^ 2-1 = 0
a=1/3
f(x)=1/3x^3-x+3

Let f (x) = ax-3, G (x) = BX ^ (- 1) + CX ^ (- 2) (a, B belong to R) and G (- 0.5) - G (1) = f (0)
(1) Try to find the relation satisfied by B and C
(2) If B = 0, the equation f (x) = g (x) has a unique solution in (0, positive infinity)
(3) If B = 1, set a = {x f (x) > G (x), and G (x)

(1)
g(-0.5)=-2b+4c
g(1) = b+ c
f(0) = -3
Then - 3B + 3C = - 3, that is, B-C = 1
(2) B = 0, then C = - 1
g(x)=-x^(-2)
ax-3=-x^(-2)
From the image, we can find that when A0, Z (x) = ax ^ 3-3x ^ 2 + 1
The first derivative of Z (x) is 3ax ^ 2-6x and the second derivative is 6ax-6
If the first derivative is zero, x = 0, or x = 2 / A
When x = 0, the second derivative 0, Z (x) is the minimum
The condition is satisfied only when the minimum value of Z (x) is zero
That is Z (2 / a) = 0, the solution is a = 2
So the value range of a is (negative infinity, 0] u a = 2
(3) B = 1, then C = 0
g(x)=x^(-1)
If G (x) x ^ (- 1) where x0 has x0, y (x) = 0 has two roots, one positive and one negative
Just X

Given the function f (x) = ax ^ 5 + BX ^ 3 + CX-1, if f (- 3) = 5, then f (3)=

f(x)=ax^5 +bx^3+cx-1
f(-x)=-ax^5 -bx^3-cx-1
=-(ax^5 +bx^3+cx-1)-2
=-f(x)-2
f(-3)=-f(3)-2=-5-2=-7

4 + 8 + 16 + 32 + 64 + 128? Simple calculation

4+8+16+32+64+128
=(4+16)+(8+64+128)+32
=20+200+32
=252

In the equilateral triangle ABC, the points D.E. are on the sides BC and AC respectively, and | BD | = 1 / 3 | BC |, | CE | = 1 / 3 | Ca |, ad, be
Fight by yourself! I can understand it! It's better to do more!

-And it's the same thing in the same way as the same thing in CE / Ca (the same thing is also the same thing in CE / CA), and it's also the same thing in the same way in the same way in the same way in the same way as the same case in the case of the case of the case of the case, and it's the same thing in the same way as the same thing in the same way as the same thing in the case of the case of the case of the case of a / 3, and it is the same as the case in the case of the same case of the same case as the same case in the same case as the same as the same as the same as the case of the case of the same as the same as the same as the same as the case in the case of the same as the same as the case in the case of the same as the case of the case of Ca [[Ca [[Ca [Ca] and the case [Ca [Ca] and the case] in this is the case in this is the case in the case of the case of the case of the case of I

Let vector group A1, A2, A3 be linearly independent. It is proved that vector group A1 + A2 + a3, A2 + a3, A3 are also linearly independent

Suppose that a1 + A2 + a3, A2 + a3, A3 are linearly related, then K1 (a1 + A2 + a3) + K2 (A2 + a3) + k3a3 = 0, where K1, K2, K3 are not all 0. Simplify to k1a1 + (K1 + K2) A2 + (K1 + K2 + K3) A3 = 0. Since vector group A1, A2, A3 are linearly independent, K1 = 0, K1 + K2 = 0, K1 + K2 + K3 = 0, then K1 = 0, K2 = 0, K3 =

If the two diagonals of the trapezoid are 10 and 12 in length and 6 in height, the area of the trapezoid is?

Trapezoid ABCD, ad ∥ BC, cross D point as parallel line of AC, cross the extension line of BC at e point, then quadrilateral aced is parallelogram, ∥ AC = De, ad = CE, ∥ ad ∥ BC, ∥ ADB area = △ ADC area = △ ECD area, ∥ trapezoid ABCD area = △ DBE area, cross D point as vertical line of be, vertical foot is H point, then in right angle △ BDH, BD = 10, DH = 6, ∥ BH = 8, similarly, in right angle △ DHE, de = AC = 12, DH = 6, ∥ eh = 6 ∥ 3, The results show that the area of Δ DBE = &# 189; be · DH = &# 189; × (8 + 6 √ 3) × 6 = 24 + 18 √ 3 = trapezoidal area

Given the quadratic function y = a (X-H) * 2, when x = 2, there is a maximum value, and the image of this function passes through the point (1, - 3)

According to the meaning of the problem, when x = 2, there is a maximum, so h = 2 and a < 0, and the image of this function passes through points (1, - 3), so: - 3 = a (1-2) ^ 2, the solution is: a = - 3, so the analytic expression of this function is: y = - 3 (X-2) ^ 2

A cone-shaped grain pile is 1.2m high and covers an area of 15m2. If you put this pile of grain into the granary, it just occupies 356 of the granary space. What is the volume of the granary?

13 × 15 × 1.2 △ 356, = 6 △ 356, = 6 × 563, = 112 (cubic meters); answer: the volume of this granary is 112m3

If a chord passes through a certain point (1,0) in the ellipse X29 + y24 = 1, then the trajectory equation of the midpoint of the chord is______ ．

Let the coordinates of the two ends of the chord be (x1, Y1), (x2.y2) and the coordinates of the middle points of the chords be (x, y). The slope of the straight line where the chord is located is kx219 + y214 = 1x229 + y224 = 1. By subtracting the two formulas, 19 (x1 + x2) (x1-x2) + 14 (Y1 + Y2) (y1-y2) = 0, that is, 2x9 + 2y4k = & nbsp; 0 and ∵ k = YX − 1

Let f (x) = ax-3, G (x) = BX ^ (- 1) + CX ^ (- 2) (a, B belong to R) and G (- 0.5) - G (1) = f (0) (1) Try to find the relation satisfied by B and C (2) If B = 0, the equation f (x) = g (x) has a unique solution in (0, positive infinity) (3) If B = 1, set a = {x f (x) > G (x), and G (x)