High score! Urgent. Senior 1 math exponential function

1.Let t = 3 + 2 x - x 2

3+2x-x 2>0 solves -11

then ax-x 2 must be incremented, so a 2> = 4

a>=8

The simultaneous solution is a>=8

So to sum up.

I hope you are satisfied, thank you.

1.Find the monotonic interval and range of y=log(base 1 2) (true number 3+2x-x 2).

The function g(x)=-x 2+2x+3 has a symmetry axis of x=1, -x 2+2x+3>0 to get -10, and the definition domain is (0,a), so g(x) increases at (0,a 2] and [a 2,a) decreases.

When 04 has no solution.

When a>1.

then a 2>4

Obtain a>8

The exchange method is a golden key to high school mathematics.

1) Let u=3+2x-x 2 (u is greater than 0), then x is greater than -1 and less than or equal to 3

Substitute to get u greater than 0 and less than or equal to 4

So true number = u is greater than 0 and less than or equal to 4 in (-1,1] increment, [1,3) decreases.

Draw the image of the log with a base of less than 1

Finally, (-1,1] decreases, [1,3) increases.

The value range is greater than or equal to -2

Let u=ax-x 2 (u greater than 0).

Then in [2,4], a is greater than x, that is, a is greater than 4 constantly.

So log is first and foremost an increment function.

Let's find you again.

u=ax-x 2 (u greater than 0).

x-a 2) 2+a 2 4 (a greater than 4) (simplify first) draw the graph (-(x-a 2) 2+a 2 4) (define the domain a greater than 4) and then draw the log diagram.

To increment the log, u should be incremented so the range of -(x-a 2) 2+a 2 4 (a greater than 4) should be incremented (as seen on the graph).

So a 2 is greater than 4

A is greater than 8 haven't been counted for a long time.,If you have any questions, find the teacher.,The teacher is very powerful (seckill!) ）

1, first of all, log(base 1 2) is a subtraction function. Let f(x)=3+2x-x 2, then f(x)=-(x-1) 2+4

Then f(x)>=0, there is a range of values of x in f(x) (-1,3), f(x) increases at (-1,1) and decreases at (1,3), then the member function decreases at (-1,1) and increases at (1,3). The value range of y is [-2, infinity).

2. First consider a>0, and a is not equal to 1

Consider the axis of symmetry to be a 2, which is certainly on the side of [2,4], Classification Discussion, >4, a>8

So the log function is an increment function so that f(x)=ax-x 2, then there is f(2)=2a-2>0, there is a>1, and a>8 is obtained

2,a<4, where f(x) is a subtraction function, so the log function is also a subtraction, i.e., 00,a<4So 08

It may be the result of the above, oral calculation. Hehe.

1.3+2x-x 2>0 solves -10,01 and vice versa.

Question 1 (-1,1> minus (1,3) increase.

2, positive infinity) should be right.

1.The base number is 1 2<1, so only the monotonicity of the true number in the range of 0 is required, that is, (x-3) (x+1) < 0, and -1 is obtainedSince the axis of symmetry of < x<3 is x=1, the monotonic increase interval of the function is [1,3) and the subtraction interval is (-1,1], and the range is [-2,+ < p> due to (3+2x-x 2) (0,4).

1) From the inscription, the inequality is 4 x<3 x, and both sides of the inequality are positive, (3 4) x<1

x 02) so that 2 x=t (I take what you wrote in parentheses) is inequality to t-1 t-1 5 3

Simplification has 3t 2-8t-3 0

Solve t 3 or t -3 5 (rounded).

i.e. 2 x 3, x log2(3).

1.Both the square of x + the square of x less than x - mx + m+4 is constant, and the side ruler solution is m=-1

2.In negative infinity to 0, monotonic panicle height decreases, and in 0 to positive infinity monotonically increases, where there is a closed interval for family bi 0.

3.When a is greater than zero, less than 1 x less than -6 7, when a is greater than 1 x greater than -6 7

Go to Counseling Education on the 8th floor of the Friendship Building.

There is no solution to the first question.

If the question is not wrong, you can start with.

The paradox of equation (x-2) multiplied by (x-squared-4)=(2-x) multiplied by x+2 simplify: (x-2) multiplied by (x+2)=-1 If the problem is wrong, then I think we should not ask the range of x values but x=?

Because, the title gives an equation with exact roots.

Question 2: 131 31

Obtained from conditional simplification.

x-2*√xy-15*y=0

Then divide both sides by x

Get the equation about (y x) and find (y x)=1 5 or -1 3 (rounded).

Bring in the required formula and get:

Original = 131 31

1.To make the equation hold, first x 2 (the non-negative requirement of the radical), the classification is discussed:

When x=2, the equation holds.

When x>2, we get the contradictory formula of (x+2)=-1, so as to sum up: x=2

2. The upstairs method is right

1.Solution: The original equation i.e.

3^(x^2)<3^(2-x).

So x 2<2-x

i.e. x 2+x-2<0

-20, a 2-3) x>(2a) x

So a 2-3>2a>0, and 2a is not equal to 1

i.e. a 2-2a-3>0, and a >0, a does not equal 1 2

Solution a>3

Instructions:1Question 1 uses the monotonicity of the exponential function to get x 2<2-x

Then solve the quadratic inequality of one element.

2.Question 2 makes use of different exponential function images.

For y=a x and y=b x, (a>b>0 and a,b is not equal to 1).

When x>0, a x must be above b x.

This can be verified by images 2 x, 3 x, (1 2) x, (1 3) x.

3.Note that in the exponential function y=a x, a>0 and a is not equal to 1

Solution: Let 2 (x+1)=a.

Then 4 (x+1)-2 (2+x)-24=[2 (x+1)] 2-2 2 (x+1)-24

a^2-2a-24

a-1)^2-25

a-6)(a+4)

2 (x+1)-6][2 (x+1)+4] doesn't seem to be able to be simplified anymore.