On the problem of function, ask the predecessors to answer: if f (x) = 3x + 7, then f (1 / x) = 3 (1 / x) + 7, why does it become like this

On the problem of function, ask the predecessors to answer: if f (x) = 3x + 7, then f (1 / x) = 3 (1 / x) + 7, why does it become like this

If f (x) = 3x + 7, then f (T) = 3T + 7 (x = t) f (a) = 3A + 7 (x = a) f (1) = 3 + 7 = 10 (x = 1), no matter what x is, you just need to bring it in
I don't know. It's like taking 1 / X as X and substituting it into the first one
For example, f (X & # 178;) = 3x & # 178; + 7
If f (2x + 1) = 3x + 2 and f (a) = 4, then a=______ .
Let 2x + 1 = a, then x = a − 12, so f (a) = 3A − 32 + 2  3a − 32 + 2 = 4, the solution is a = 73, so the answer is 73
Let f (2x + 1) = 3x-2, then the analytic expression of F (x) is-------
f(2x+1)=3x-2=1.5*(2x+1)-3.5
f(x)=(3x-7)/2
Then the equation of degree m x + m is known?
M is not equal to - 3
The inequality ^ 2 (log + 2) ^ X-7 has the maximum solution
Y = loga ^ LG (x ^ 2-2x + 3) has the maximum value, and the solution to the inequality loga (x ^ 2-5x + 7) > 0
A>1
x^2-5x+7>1
(x-2)(x-3)>0
x> 3 or X
Is log based on a LG (x ^ 2-2x + 3) true?
If yes, then x ^ 2-2x + 3 = (x-1) ^ 2 + 2
Then LG (x ^ 2-2x + 3) = LG [(x-1) ^ 2 + 2] ≥ LG2 > 0
Because there is a maximum, so 0
If the solution of the system of equations 3x + y = 4K + 3 x + 3Y = K + 1 satisfies - 2 less than X-Y less than or equal to 3, the value range of K can be obtained
1. Find out the solution of the equations x, y, x = 11K / 8 + 1, y = - K / 8
2. Solving the system of inequalities - 2
-7/9〈k〈8/9
Triangle area and perimeter formula, circle area and perimeter formula
Triangle: S = 1 / 2ah or S = 1 / 2absinc, C = a + B + C
Circle: S = π R ^ 2 C = 2 π R
Divide the area of a triangle by the height of its base by its two circumferences: the sum of its three sides
If 2 is the solution of the equation 3x + 4A = x / 2-A about X, then a ^ 2013 - (- a)=
Wrong. If - 2 is the solution of the equation 3x + 4A = x / 2-A about X, then a ^ 2013 - (- a)=
Substituting x = - 2 into the equation yields
-6+4a=-1-a
∴a=-1
∴a^2013-(-a)=-1-1=-2
If the equation (m ^ 21) x ^ 2-mx + 8 = x is a unary linear equation about X, then the value of the algebraic formula m ^ 2008 - | M-1 | is ()
Title, such as
On the linear equation with one variable of X
There are no quadratic terms
So the quadratic coefficient is 0
m^2-1=0
m=±1
The coefficient of the first term is not 0, M is not equal to 0
Here
If M = 1, m ^ 2008 - | M-1 | = 1-0 = 1
If M = - 1, m ^ 2008 - | M-1 | = 1-2 = - 1
Let f (x) satisfy f (x ^ 2-3) = loga (x ^ 2) / (6-x ^ 2) (a > 0 and a ≠ 1)
It is proved that the function f (x) is monotonically increasing in its domain of definition when a > 1
Proof: let t = x ^ 2-3, then: x = ± √ T + 3. So: F (T) = loga [(T + 3) / (6-t-3)] = loga [(3 + T) / (3-T)] that is, f (x) = loga [(3 + x) / (3-x)] let x1, X2; if X1 > X2, then f (x1) - f (x2) = loga [(3 + x1) / (3-x1)] - loga [(3 + x2) / (3-x2)] = loga {[(3 + x1) / (3-x1)] / [(3 + x2) / (3 -