Y = square 2 (x) of log4 (x-1)

Y = square 2 (x) of log4 (x-1)

(x-1)^2=4^y=2^2y=(2^y)^2
x-1

Inverse function of x square minus 1 of y = 2
seek

X power of 2 = y + 1
X = logarithm of (y + 1) with base 2
So the inverse function is y = logarithm of (x + 1) with base 2

The inverse of the function y = x / 1 + X is

y=x/(x+1)
xy+y=x
xy-x=-y
x-xy=y
x(1-y)=y
x=y/(1-y)
f(-1)(x)=x/(1-x).

Given the two real roots of proposition p: X1 and the square of x-mx-2 = 0, the square of inequality a-5a-3 is greater than or equal to [x1-x2] for any real number m? [- 1,1]
Given the two real roots of proposition p: X1 and the square of x-mx-2 = 0, the square of inequality a-5a-3 is greater than or equal to [x1-x2] for any real number m? [- 1,1]; proposition q: the square of inequality ax + 2x-1 > 0 has a solution. If proposition p is a true proposition and proposition q is a false proposition, find the value range of A

x1+x2=m,x1*x2=-2
|x1-x2|=√[(x1+x2)^2-4x1x2]=√(m^2+8)
The maximum value of F (m) in M ∈ [- 1,1] = √ 9 = 3
a^2-5a-3≥3
(a+1)(a-6)≥0
A ≥ 6 or a ≤ - 1
Ax ^ 2 + 2x-1 > 0 has no solution, ax ^ 2 + 2x-1 ≤ 0
a

Draw the function image of (1) y = - 2-1 and y = - 2-3, and talk about their relationship? (2) y = 2x + 3 and y = 2x-3
Images and answers

1. Y = - 2-1 = - 3, y = - 2-3 = - 5. The images of these two functions are below the x-axis, parallel to the x-axis, and the distances from the x-axis are 3 and 5, respectively. I guess the problem is wrong, not so simple. The image of 2, y = 2x + 3 is a straight line passing through two points (0,3) and (- 3 / 2,0); the image of y = 2x-3 is a straight line passing through two points (0, - 3)

The vertex coordinates of the image of the quadratic function y = 2x & # - 8x are

Solution
y=2x²-8x
=2(x²-4x)
=2(x²-4x+4)-8
=2(x-2)²-8
The vertex coordinates are (2, - 8)

Vertex coordinates of quadratic function y = - 2x & # 178; - 6x + 1

y=-2x²-6x+1
=-2(x^2+3x)+1
=-2(x^2+3x+9/4)+9/2+1
=-2(x+3/2)^2+11/2
Vertex coordinates (- 3 / 2, 11 / 2)

If the maximum value of quadratic function y = ax ^ 2-4x + A-3 is always negative, how to find the value range of a?

There is a maximum value, indicating that the opening is downward, a

If the maximum value of quadratic function y = MX ^ 2-4x + M-3 is negative, then the value range of M is

First, if the function has a maximum, then M0;
m> 4 or M

The maximum value of quadratic function y = 2x ^ 2 + 4x-9 is
A.9 B.-9 C.-7 D.7

y=2x^2+4x-9
=2(x+1)^2-11
X = - 1, y min - 11
There is no maximum
If:
y=-2x^2+4x-9
=-2(x-1)^2-7
X = 1, the maximum is - 7
Choose C. - 7