My son is 12 years old and my father is 39 years old A. Three years later B. three years ago C. nine years later D. impossible

After X years, the father's age is four times that of his son. According to the meaning of the question: 39 + x = 4 (12 + x), the solution is: x = - 3, that is, three years ago, the father's age was four times that of his son

Finding the range of X-2 under the root sign of y = 3x-6

Y = 3x-6 √ (X-2) let √ (X-2) = t, t ≥ 0, x = t + 2 ≥ 2, y = 3 (T + 2) - 6T = 3t-6t + 6 = 3 (t-1) + 3. When t = 1, i.e. x = 3, y has the minimum value of 3. The range of value is [3, + ∞]

This year, my father is 40 years old, and my son is 12 years old. A few years later, my father's age is just three times that of my son?
Don't set X

Digital solutions:
Difference between father and son: 40-12 = 28 years old
Son's present age: 28 / (3-1) = 14 (years old)
Years later: 14-12 = 2 (years)

What is the human range of the function y = (1-x quadratic) parts (1 plus x quadratic)

Y=(1-X²)/(1+X²)
=［2-(1+X²)］/(1+X²)
=2/(1+X²)-1
∵1+X²≥1,
∴Y≤2-1=1,
Range: (- ∞, 1)

My father is 50 years old and my son is 14 years old. A few years later, my father will be three times as old as my son?

Suppose that the age of the father after X years is three times that of the son, and the meaning of the question is: 3 (x + 14) = 50 + X & nbsp; & nbsp; & nbsp; 3x + 42 = 50 + X & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; 2x = 8 & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; & nbsp; X = 4 A: the age of the father after 4 years is the age of the son

The range of the function y = 3-x squared / 2x squared + 5 is

∵2x²＋5＞0
∴x∈R

The father is 39 years old, and the son is 11 years old. A few years later, the father's age is exactly three times that of his son (solved by equation)

Suppose that after X years, the father's age is exactly three times that of his son
39+X=3*(11+X)
39+X=33+3X
2X = 6
X=3

The monotone decreasing or monotone increasing of F (x) on D exists in the interval [a, b]. The range of F (x) on [a, b] is [a, b]
(1) Find the interval of the closed function f (x) = x ^ 3 meeting condition 2
(2) Judge whether the function y = 2x + lgx is a closed function, and explain the reason
(3) If the function y = K + radical (x + 2) is a closed function, find the value range of K
How to solve such a problem? I don't know how to start

(1) , f (x) = x ^ 3 is monotonically increasing on R, so the range of interval [a, b] is [a ^ 3, B ^ 3],
A = a ^ 3, B = B ^ 3. So a = 0, B = 1
So the interval is [0,1]
(2) . y '= 2 + 1 / x, x > 0. So y' = 0
Let the interval [a, b] be any interval in its domain
f(a)=2a+lga.,f(b)=2b+lgb
Then a = -- LGA
10^a=1/a
10 ^ A * a = 1. The equation has a solution, so it is a closed function
(3) Function is an increasing function. So
X = K + radical (x + 2)
（x-k)^2=x+2
x^2-(2k+1)x+k^2-2=0
For a function to be closed, the equation must have a solution
So: (2k + 1) ^ 2-4 (k ^ 2-2) > = 0
4k^2+4k+1-4k^2+8>=0
4k>=-9
So: k > = -- 9 / 4

My son is 11 years old and my father is 39 years old. How many years later, my father is twice as old as his son

17 years later, the father is twice as old as his son

Given the first-order function y = KX + B, when 0 ≤ x ≤ 2, the range of corresponding function value y is - 2 ≤ y ≤ 4, try to find the value of KB

(1) When k > 0, y increases with the increase of X, that is, the first-order function is an increasing function, when x = 0, y = - 2, when x = 2, y = 4, substituting the analytic formula of first-order function y = KX + B to get: B = − 22K + B = 4, the solution is k = 3B = − 2, ｜ KB = 3 × (- 2) = - 6; (2) when k < 0, y decreases with the increase of X, that is, the first-order function is a decreasing function, when x = 0, y = 4, when x = 2, y = - 2, substituting first-order function The analytic formula of function y = KX + B is: B = 42K + B = - 2, the solution is k = - 3B = 4 KB = - 3 × 4 = - 12. So the value of KB is - 6 or - 12

My son is 12 years old and my father is 39 years old A. Three years later B. three years ago C. nine years later D. impossible