Senior 1 math questions, urgent answers, math senior 1 questions for answers

Solution: According to the topic.

It can be solved by the difference method.

Let y=x 3-x-1

When x=0, y=-1

When x=2, y=5

Insert x3=1 y=-1

Insert x4= y= * 1 >0

Insert x5=5 4 y=125 64 -5 4-1 <0 x5=insert x6= 11 8 y= 1331 512 -11 8-1= 819 512 -11 8 =115 512 >0 x6=

Insert x7=21 16 y = <0 x7=insert x8=43 32 y= 26>0 x8=insert x9=85 64 y=>0 x9=

So x should be between x7 and x9, and because it only needs to be accurate, the solution is.

Let y=x 3-x-1

Derivative first, get y=x 2-1

Let y=0 give x=3 3 and 3 3

The original function y is in — 3 3 is the increment function.

At 3 3, 3 3 is a subtraction function.

In 3 3, is the increment function.

When x=0, y=-1

When x=1, y=-1

When x=2, y=5

So the function has only one solution between (0,2).

Use the dichotomy to find it.

x= y= * 1 >0

x=5/4 y=125/64 -5/4-1 <0x= 11/8 y= 1331/512 -11/8-1= 819/512 -11/8 =115/512 >0

x=21/16 y = <0 x7=

x=43/32 y= 26>0 x8=

x=85/64 y=>0 x9=

The solution is in between. It has to be interpreted as.

You can extract the unknown number x and then put the first and last 2 generations in the interval. Calculated (,

Generally, interval questions are answered by substituting the first and last 2 numbers of the interval into the equation. Although sometimes pay attention to the addition and subtraction of the equation.

Use a dichotomy! Got it?? It's Newton's dichotomy! It's simple! It's also in the book.

Analysis: First of all, the title said that "if the water consumption does not exceed 5 tons, the water fee per ton is yuan, if it exceeds 5 tons but does not exceed 6 tons, the excess part of the water fee will be charged at 200% of the original price, and if it exceeds 6 tons and does not exceed 7 tons, the excess part of the water fee will be charged at 400% of the original price", and said "x(x<=7) tons" So all three situations should be discussed.

Answer: Distribution function.

x* (x<=5)

5* (5x<=6)

5* (6

Because his water consumption does not exceed 7 tons, it is discussed in three cases, namely:

1: (x<=5)

2: (RMB) (53: (RMB) (6

When 6 is 5 when x< = 5, y=x*

This is a piecewise function.

Since f(x) is a quadratic function, let f(x)=ax +bx+cFirst, f(x)+g(x) is an odd function, let this odd function be t(x) so t(0)=0, and g(x)=-x -3

Substituting t(0)=f(0)+g(0)=c-3=0 c=3 f(x)=ax +bx+3

The odd function t(x) has t(1)+t(-1)=0

Substituting yields: t(1)+t(-1)=f(1)+g(1)+f(-1)+g(-1).

a+b+3-4+a-b+3-4

2a-20 a=1 f(x)=x +bx+3 The image opening is upward, and the axis of symmetry is x=-b 2

Discussion in conjunction with image classification).

The axis of symmetry is to the left of -1, that is, when x=-b 2 -1, the b 2 image is obtained when x [-1,2] is minimum x=-1, and substituting f(-1)=1-b+3=1, b=3 2, is true;

When the axis of symmetry is between [-1,2], it is smallest at -1 -b 2 2 b -4 image x = -b 2.

Substitute f(-b 2) = b 4 - b 2 + 3 = -b 4 + 3 = 1 b = 2 2 ( 2 root number 2).

and 2 b -4, 2 2 2, rounded, -2 2 conformed, established;

The axis of symmetry is on the right side of 2, that is, when the edge x=-b 2 2, the b-4 image is obtained when x [-1,2] is minimum x=2, and is substituted for f(2)=4+2b+3=1b=-3 -4, and rounded.

In summary, the value of b is 3 or -2 2.

So f(x)=x +3x+3 or f(x)=x -2 2x+3.

Do you dare to add some points, it's so difficult!

Solution: From the problem, f(1+y 2)=3y+5y-1 makes 1+y 2=2x-1 then y=4x-4 substitution.

obtained: f(2x-1)=48x-76x+272f(x)+f(1 x)=2x.1

2f（1/x）+f（x)=2/x ..2

Synoptic f(x) = 4x 3-2 (3x).

Hope it works for you!

It is added that f(5)=f(3+2)=-1 f(3), f(1+2)=-1 f(1), so f(1)=f(5)=-5, it can be seen that the function is a function with a period of 4.

f(f(1))=f(-5)=f(3)=-1/f(1)=1/5

1) Let t=1+x2, then x=2t-2

f(t)=3(2t-2)²+5(2t-2)-112t²-14t+1

f(2x-1)=12(2x-1)²-14(2x-1)+148x²-76²+27

2) Replace x in a known equation with 1 x to obtain.

2f(1/x)+f(x)=2/x

f(x) and f(1 x) are regarded as unknowns, and they are solved in conjunction with the known equations.

f(x)=(4/3)x-2/(3x)

x²-100x+49

I'm substituting 1+x 2 with x as a whole, then replacing 2x-1 with an equation containing x, and then solving it)

2. There is nothing that can be done. Sweat...

Vector m1m2 = (-5,5).

Let the intersection coordinates of the line m1m2 and the line y=mx-7 p(x,y) vector m1p vector m1m2=3 5, that is, x-6=(3 5)*(5) get x=3, y-2=(3 5)*5 get y=5

Therefore, p = (3, 5) substituting p point into the straight line y=mx-7 to solve m=4 method 2:

Score point: Let the intersection point of the straight line and ab be p(x,y).

There is x=(x1 + x2) (1 + =(6+3 2 1) (1+3 2)=3

y=(y1 + y2) / (1 + =(2+3/2 ×7)/(1+3/2)=5

p(3,5) is substituted into y=mx-7, and m=4[ indicates the ratio from the starting point to p to the end point from p to the end point].

How do I solve the equation for the line where the intersection coordinates are (3,5) or (-9,17) and the line segment is y=-x+8

The line where the intersection is located is also on the line y=mx-7, so the solution is m=4 or m=-8 3

a∈[0,8]

Since the domain is defined as r

Let g=ax 2-ax+2 for x r always set up a classification discussion.

a 0 quadratic function with the image opening up so that the image is all above the x-axis, i.e. 0 gets 0 a 8

a=0 and g=2 are greater than 0 when any value is taken x, so a=0 is true, and when a0 is established, the opening is downward, and it can be seen that the quadratic function image must have a part below the x-axis, so a<0 is not true.

All of the above are taken and combined to obtain a [0,8].

Classify. When a=0 holds. When a is less than 0, the opening is downward. Drawing knows not satisfied.

Delta = a squared - 4a * 2 less than or equal to 0It is concluded that a is less than or equal to 8 and greater than or equal to 0. The so-called definition domain is equal to r, that is, any can be, then the root number under the constant is greater than equal to 0, that is, the image should not be below the x-axis.