The definition domain of the inverse function of x ^ 2-2x + 3 (x ≤ 1) under y = root sign is?

The definition domain of the inverse function of x ^ 2-2x + 3 (x ≤ 1) under y = root sign is?

The definition domain of the inverse function of a function is the value range of the function
Y = radical (x ^ 2-2x + 3)
=Radical [(x-1) ^ 2 + 2]
The minimum value is root 2
There is no maximum,
Therefore, the value range of this function is: [radical 2, + ∞)
Therefore, the definition domain of inverse function is: [radical 2, + ∞)

The square of 2x plus 3x equals 3

2x² + 3x = 3
2x² + 3x - 3 = 0
X = (- 3 + √ 33) / 4 or x = (- 3 - √ 33) / 4

Y = big {above is negative x, X is greater than or equal to minus one, and below is the square of X

Y = big {above is - x, X is less than or equal to 1, below is negative root sign, X is greater than 1

How to solve 3 / 5-4x square + 3x greater than or equal to (2x + 1) (1-2x)

3/5-4x²+3x≧(2x+1)(1-2x)
3/5-4x²+3x≧-4x²+1
3x≧2/5
x≧2/15

The polynomial that subtracts 2x equals x2-3x-6 is () A. x2-5x-6 B. x2+5x-6 C. x2-x-6 D. x2+5x+6

The polynomial obtained is: (x2-3x-6) + 2x
=x2-x-6．
Therefore, C

1. Y = 2sinx, X belongs to the closed interval - π / 6, π / 62, y = 1 + ln (x + 2) 3, y = 2 x power / 2 inverse function Can you give me a detailed answer to the three questions of the inverse function used in the knowledge points

1. Y = 2sinx, X belongs to the closed interval - π / 6, π / 6
First determine the value field of the function, which is also the definition field of the inverse function
y = 2sin(-π/6) = -1
y = 2sin(π/6) = 1
y=2sinx
sinx = y/2
x = arcsin(y/2)
The inverse function is
Y = arcsin (x / 2), X belongs to closed interval [- 1,1]
2、y=1+ln(x+2)
y - 1 =ln(x+2)
x + 2 = e ^(y-1)
x = e ^(y-1) -2
The inverse function obtained is
y = e ^(x-1) -2
3、y =2^x / 2^(x+1)
Where should the brackets be? I'm not sure. Please make the title clear

Let f (x) = ln (x + 1), and let the inverse function of F (x) be f '(x) Function problem: given f (x) = ln (x + 1), let the inverse function of F (x) be f '(x) 1. Find the monotone interval of G (x) = f (x) - f '(x). 2. For any x > 0, the inequality LNF' (x) - f (e to the power of x) is obtained

1. Get the derivative and make it equal to 0:1 / (x + 1) - e ^ x = 0. Draw the curve of 1 / (x + 1) and e ^ x, and you can find that the intersection point is (0,1). Both sides of the intersection point are monotone intervals. Ah (^ means factorial)
The second one is not. I haven't cared about Mathematics for a long time. But the result given by Matlab is: a < 2.02

Let y = f (x) be an odd function. When x ≥ 0, f (x) = 3x-1. Let the inverse function of F (x) be y = g (x), then G (- 8)=______

Method 1: when x < 0, - x > 0, the known f (- x) = 3-x-1
And ∵ f (x) is an odd function,
ν f (- x) = - f (x), i.e. - f (x) = 3-x-1
∴f（x）=1-3-x．
∴f（x）=
3x−1
1−3−x
x≥0
x<0.
∴f-1（x）=
log3(x+1)      x≥0
−log3(1−x)    x<0.
∴f-1（-8）=g（-8）=-log3（1+8）=-log332=-2．
Method 2: when x < 0, - x > 0, the known f (- x) = 3-x-1
And ∵ f (x) is an odd function,
ν f (- x) = - f (x), i.e. - f (x) = 3-x-1
ν f (x) = 1-3-x
Let 1-3-x = - 8 get x = - 2, that is: G (- 8) = - 2
The answer is: - 2

Find the inverse function of F (x) = X-2 / 3x + 4

y=(x-2)/(3x=4)
3xy+4y=x-2
(3y-1)x=-4y-2
x=(-4y-2)/(3y-1)
So the inverse function is f (x) = - (4x + 2) / (3x-1)

Inverse function of y = x / (3x + 5) How to find the definition domain of inverse function of y = x / (3x + 5)

First of all, the range of values obtained is a part of the definition domain, and then do the inverse function to see whether y can be 0. If not, it can be supplemented at last