Is 5 / 2 of y = x (x is an independent variable) an inverse proportional function? Is 1 / 5 of y = x (x is an independent variable) a positive proportional function

Is 5 / 2 of y = x (x is an independent variable) an inverse proportional function? Is 1 / 5 of y = x (x is an independent variable) a positive proportional function

analysis
y=5/x²
Not an inverse scale function
y=1/5x
It's a positive scale function

Given the relation y = KX + 2, and the independent variable x = - 3, the dependent variable y = 0, then when the independent variable x = 9, the value of the dependent variable y is 0______ ．

From the meaning of the question, - 3K + 2 = 0, the solution is k = 23, so the relation is y = 23x + 2, when x = 9, y = 23 × 9 + 2 = 6 + 2 = 8

Is the inverse proportion function a function of one degree, a function of several degrees, an independent variable X or a dependent variable y?

First, the analytic form of the function is integral, that is, the independent variable will not be in the root or denominator. Second, the highest degree of the independent variable is 1, which is called the first-order function. The highest degree of the independent variable is 2, which is called the second-order function, It does not belong to the category of linear function

Please ask again, does Hanken's function usually need to be solved after square?

If y = ax + radical (BX + C), we usually use the substitution method,
Let t = root (BX + C)
If there are two radicals and the coefficients of X are opposite, the square is better
If the root sign (a ^ 2 - x ^ 2), the substitution method X = acost
etc.

It is known that the simplest radical x + y + 1 and root 3x-y + 11 can be combined into one term to find the value of root X-Y

Can be combined into one
Then x + y + 1 = 3x-y + 11
2x-2y=-10
x-y=-5
So the root X-Y is meaningless

Given that the function f (x) = 2|x + 1| + ax a belongs to R, it is proved that f (x) is an increasing function on R when a > 2

1. When x + 1 > = 0, x > = - 1, f (x) = 2 (x + 1) + AX = (2 + a) x + 2, let x2 > x1f (x2) - f (x1) = (2 + a) x2 + 2 - (2 + a) x1-2 = (2 + a) (x2-x1) when a > 2, 2 + a > 0, x2-x1 > 0f (x2) - f (x1) > 0, so when x > = - 1, a > 2, f (x) is increasing function 2, when x + 12, A-2 > 0, x2-x1 > 0f (x2) - f (x1) > 0, so

Prove the increasing function of function f (x) = ax + 1 (a > 0) on R

Let x2 > x1
f(x2)-f(x1) = (ax2+1)-(ax1+1) = a(x2-x1)
∵x2＞x1
∴x2-x1＞0
Also: a > 0
∴ a(x2-x1)＞0
∴f(x2)＞f(x1)
The increasing function of F (x) = ax + 1 (a > 0) on R

1. Let a > 0, f (x) = x + X / A, and f (- 1) = - 5 (1) find the value of a (2) prove: F (- 1) + F (x) = 0
2. Let u = R, f (x) = root (1-x) + root (x + 2) be defined as a, and G (x) = x-2x-3 be defined as B
The first question is wrong
Let a > 0 function f (x) = x + A ^ 2 / X

1
f(-1)=-1-a²=-5,∴a=±2
And ∵ a > 0, ∵ a = 2
∴f(x)=x+4/x
(2) Certification:
f(-x)+f(x)=-x-4/x+x+4/x=0
2. Solutions: a = {x | 1-x ≥ 0, x + 2 ≥ 0} = {x | - 2 ≤ x ≤ 1}, ∧ CUA = {x | x ＜ - 2 or X ＞ 1}
B={y|y=x²-2x-3=(x-1)²-4}={y|y≥-4}
∩ CUA ∩ B = {x | - 4 ≤ x ＜ 2 or X ＞ 1}

It is proved by complete inductive reasoning that the function f (x) = octave power of X - quintic power of X + quadratic power of X - x + 1

F (1) = a + B = - 1, f (2) = 4A + B = 8, a = 3, B = - 4, f (x) = 3x ^ 2-4, f (5) = 71, f (1) = - 1, then a + B = - 1, f (2) = 4, f (x) = 3x ^ 2-4, f (5) = 71, f (1) = - 1（

Find the monotone interval of function f (x) = 2 / 3 power of X + 1 / 3 power of X and prove your conclusion

f(x)=[(x)^(1/3)]^2+(x)^(1/3)+1
Let (x) ^ (1 / 3) = X
f(X)=X^2+X+1=(X+1/2)^2+3/4
When x = - 1 / 2, f (x) = 0
X