The image of function y = f (x) and G (x) = (x-1) ^ 2 (x) A：f（x）=1+√x（x>=0） B：f（x）=1-√x（x>=0）

The image of function y = f (x) and G (x) = (x-1) ^ 2 (x) A：f（x）=1+√x（x>=0） B：f（x）=1-√x（x>=0）

The image of function y = f (x) and G (x) = (x-1) ^ 2 (x = 0, range f (x)

F (x) = the third power of X + X + 1 (x belongs to R) what is the symmetry of an image about a point

Because y = x ^ 3 + X is symmetric with respect to (0,0),
therefore
F (x) = the third power of X + X + 1 about (0,1) symmetry

If the image of power function f (x) is over (2,8), then f (x)=

Let f (x) = x ^ n
∵f(2)=2^n=8
∴n=3
∴f(x)=x^3

There are 10 rational numbers, including 6 positive numbers and 6 integers. The number of negative fractions is equal to that of positive fractions, and the number of negative fractions is no more than 3. How many negative integers are there?
How many positive integers, negative integers, positive fractions and negative fractions are there?

There are 6 positive numbers and no more than 3 negative numbers, so one number is 0 and the other three are negative
Six integers means that there are four fractions. The number of negative fractions is equal to that of positive fractions, so there is only one negative integer

The function f (x) = ax3-4x + 4 (a ∈ R) is known to have an extreme value at x = 2. (I) determine the value of a and find the monotone interval of the function; (II) if the equation f (x) = B about X has at most two zeros, find the value range of real number B

(I) because f (x) = ax3-4x + 4 (a ∈ R), f ′ (x) = 3ax2-4. Because f (x) has an extreme value when x = 2, f ′ (2) = 0, that is, 3 × 4a-4 = 0, we get & nbsp; & nbsp; So f (x) = 13x3 − 4x + 4, so f ′ (x) = x2-4 = (x + 2) (X-2) Let f ′ (x) = 0, x = 2, or x = - 2, when x changes, f ′ (x), f (x) changes as follows: X (- ∞, - 2) - 2 (- 2, 2) 2 (2, + ∞) f ′ (x) + 0 - 0 + F (x) monotonically increases ↗ Monotonic decrease of maximum ↘ Monotonically increasing minimum ↗ So the monotone increasing interval of F (x) is (- ∞, - 2), (2, + ∞); the monotone decreasing interval of F (x) is (- 2, 2). (II) from (I), we know that when x = - 2, f (x) has a maximum, and the maximum is f (− 2) = 283; when x = 2, f (x) has a minimum, and the minimum is f (2) = − 43; if the equation f (x) = B has at most two zeros, then the value of B The range is (− ∞, − 43] ∪ [283, + ∞)

The reciprocal of three fourths and the sum of four ninths are one more than the reciprocal of a number. What is the number?

This number
=1 ÷ (3 / 4 + 4 / 9-1)
=1 ÷ (4 / 3 + 4 / 9-1)
=7 / 9
=9 out of 7

Solving inequality 8 (Y -- 1) less than 4Y + 20 less than 8y

8（y--1）8y-8
4y-8y>-8-20
-4y>-28
y

The zero point of the function f (x) = x3-2x2 + 3x-6 in the interval [- 2,4] must be in the following interval______ within
A.[-2,1] B[5/2,4] C.[1,7/4] D.[7/4,5/2]

Substituting the two ends of the interval of each option into the function, if the values of the two functions are different, then the zero point must be in it

What is the point of intersection between the following parabola and X-axis?
A. Y = 3x square - 4x-1b. Y = x square - 2x root 2 times x-2x + 3 + 2x root 2C. Y = 10x square + 3x + 5D. Y = 2 / 2 root 3 times x square + 2x root 2 times X-1

If there is only one intersection point with the x-axis, there is only one solution to the corresponding quadratic equation of one variable
So the discriminant is equal to 0
So the discriminant in B = (- 2 √ 2-2) & sup2; - 4 (3 + 2 √ 2)
=8+8√2+4-12-8√2
=0
accord with
Choose B

Why is y ^ 2 = 5x + 25 / 4 a parabola? How to judge its axis of symmetry, vertex and opening direction?

y^2=5x+25/4
5x=y^2 -25/4
x=1/5*y^2 -5/4
Its axis of symmetry is x, its vertex is (- 5 / 4,0) and its opening direction is to the right