Piecewise function: F (x) = {2x + 3, (x)

Piecewise function: F (x) = {2x + 3, (x)

F (x) = {2x + 3, (X & lt; - 1) & nbsp;, (- 1 ≤ x < 1) & nbsp;, (- 1 ≤ x < 1) & nbsp;, X-1, (x ≥ 1)} & nbsp; (1) draw function image? (2) f (x) > 1 / 4 & nbsp;, the value range of X is? (3) f [f (x)] = 1 / 4 & nbsp;, x =?
(1) Analysis:
 
 
(2) Analysis: 2x + 3 & gt; 1 / 4 = = & gt; 2x & gt; - 11 / 4 = = & gt; - 11 / 8 & lt; X & lt; - 1
X ^ 2 & gt; 1 / 4 = = & gt; - 1 & lt; = x & lt; - 1 / 2 or 1 / 2 & lt; X & lt; 1
x-1>1/4==>x>5/4
To sum up: - 11 / 8 & lt; X & lt; - 1 / 2 or 1 / 2 & lt; X & lt; 1 or X & gt; 5 / 4
 
(3) Analysis: let t = f (x)
F(t)=1/4
The results show that T1 = - 11 / 8, T2 = - 1 / 2, T3 = 1 / 2, T4 = 5 / 4
2x+3=-11/8==>x=-35/16
X ^ 2 = - 1 / 2, no solution
X^2=1/2èx=-√2/2,x=√2/2
x-1=5/4==>x=9/4

It is known that f (x) is an odd function of the domain in the interval (- 1,1). When x belongs to (0,1), f (x) = 2x / 4x + 1
2X is the x power of 2, 4x is the x power of 4
1. Discuss the monotonicity of function f (x)
2. Find the range of function f (x)

① When x ∈ (0,1), f (x) = 2 ^ X / (4 ^ x + 1)
When x ∈ (- 1,0), then - x ∈ (0,1)
∴f(-x)=2^(-x)/(4^(-x)+1)=2^x/(4^x+1)
And the function f (x) is odd
∴f(0)=0 f(x)=-f(-x)
∴f(x)=-2^x/(4^x+1) x∈(-1,0)
f(0)=0
f(x)=2^x/(4^x+1) x∈(0,1)
∵2^x/(4^x+1)=1/(2^x+2^(-x))≤1/2
∴-2^x/(4^x+1)≥-1/2
On (- 1,0), f (x) is monotonically decreasing
It is monotone decreasing on (0,1)
That is (- 1,0) and (0,1) are monotone decreasing intervals of functions
② It can be seen from (1) that in (- 1,0) f (x) ∈ (- 1 / 2, - 2 / 5]
On (0,1), f (x) ∈ [2 / 5,1 / 2]
In conclusion, the range of F (x) is (- 1 / 2, - 2 / 5] ∪ {0} ∪ [2 / 5,1 / 2)

If a car drives from a to B, it can arrive one hour earlier than the original time if the speed is increased by one fifth. If the speed is increased by one fourth after driving 120 km at the original speed, it can arrive 40 minutes earlier?

A certain distance, speed and time are inversely proportional, 40 minutes = 2 / 3 hours
Speed after 1 / 5 increase: original speed = (1 + 1 / 5): 1 = 6:5
Time for 1 / 5 speed increase: original time = 5:6
The original time is 1 △ 6-5 × 6 = 6 hours
Increase 1 / 4 speed: original speed = (1 + 1 / 4): 1 = 5:4
Time for 1 / 4 speed increase: original time = 4:5
Increase the distance of 1 / 4 speed, the original time = 2 / 3 ÷ (5-4) × 5 = 10 / 3 hours
Driving 120 km at the original speed takes 6-10 / 3 = 8 / 3 hours
Original speed = 120 △ 8 / 3 = 45 km / h
The distance between a and B is 6 × 45 = 270 km

As shown in the figure, the two heights BD and CE of △ ABC intersect at point h. are points a, e, h and D on the same circle? Please give reasons

Points a, e, h and D are on the same circle. The reasons are as follows: connect ah, take the midpoint o of ah, connect OE and OD, as shown in the figure. ∵ the two high BD and CE of ABC intersect at point h, ∵ AEH = ∠ ADH = 90 °.∵ point O is the midpoint of ah, ∵ EO = od = Oh = OA. ∵ points a, e, h and D are on the circle with point o as the center and OA as the radius

The first car carries 30 more than the second car, and the third car carries 20 less than the second car. How many tons can each car carry

Reference answer:
Set up a third vehicle to transport x tons
x+(x+20)+(x+20+30)=910
3x=840
x=280
280+20=300 300+30=330
A: the first car carries 330 tons, the second 300 tons and the third 280 tons

If the real number a.b.c. satisfies both a-7b + 8C = 4 and 8A + 4b-c = 7, then a & sup2; - B & sup2; + C & sup2=

This problem is a-7b + 8C = 4 ① and 8A + 4b-c = 7 ②. ① * 8 - ②
When a is eliminated, C = (12b + 5) / 13 is obtained. When C is eliminated by (1 + 2 × 8), a = (12-5b) / 13 is obtained
Substituting into the algebraic expression, B is reduced to (144-120b + 25B ^ 2) / 169-b ^ 2 + (144b ^ 2 + 120b + 25) / 169, and the final result is 1

There is a batch of goods to be transported to a certain place. The owner of the goods is going to rent two types of cars from the automobile transportation company
There is a batch of goods to be transported to a certain place. The owner of the goods is going to rent two types of trucks from the automobile transportation company. The situation of renting these two kinds of trucks in the past two times is shown in the table
The first time, the second time
Number of class a freight cars (unit: 4 10)
Number of type B freight cars (unit: 6 12)
Accumulated tonnage (unit: ton) 3170
This time, we have to hire 3 A-type trucks and 5 B-type trucks of the company, just to transport the goods to a certain place. If we pay 300 yuan per ton, how much is the freight?
Quadratic equation of two variables

The load of each type a truck is x tons, and that of each type B truck is y tons
4x+6y=31
10x+12y=70
By solving the above equations, we get x = 4, y = 2.5
（3*4+5*2.5）*300=7350

Rectangular coordinate equation x & # 178; + 9y & # 178; = 1 is transformed into polar coordinate equation

From x = ρ cos θ, y = ρ sin θ
Substituting: ρ & # 178; Cos & # 178; θ + 9 ρ & # 178; sin & # 178; θ = 1

For a pile of manatee brand cement, two thirds of the total amount is used for the first time, and 20% for the second time. How many tons of 120t cement is left?

120/(1-2/3-20%)=900

As shown in the figure, the three sides of the right triangle are 3, 4, 5, EF, eg, ED, which are perpendicular to AB, BC, CA, and EF = eg = ed, then EF is calculated

Let the side length of a square bfeg be x, then s △ ABC = s △ Abe + s △ BCE + s △ CAE, 12 × 3 × 4 = 12 × ab × EF + 12 × BC × eg + 12 × Ca × ed; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 6 = 12 × 3 × x + 12 × 4 × x + 12 × 5 × x, & nbsp; & nbsp; & nbsp; & nbsp; & nbsp