If the image of exponential function f (x) = a ^ x (a > 0, a is not equal to 1) passes through points (2,9), then f (x) =?

If the image of exponential function f (x) = a ^ x (a > 0, a is not equal to 1) passes through points (2,9), then f (x) =?

a^2=9,a>0
Then a = 3
Then f (x) = 3 ^ X

It is known that there are two symmetries on the parabola y & sup2; = x with respect to the straight line L: y = K (x-1) + 1

The straight line L: y = K (x-1) + 1 passes through the point (1,1), which is on the parabola (k is obviously not 0)
Let there be such two different points a and B on the parabola
Let the coordinates of a be (T1 & sup2;, T1) and B be (T2 & sup2;, T2), where T1 & sup2; is not equal to T2 & sup2;,
Since two points are symmetric with respect to the line L: y = K (x-1) + 1, then
（t1-t2)/(t1²-t2²)=-1/k,
K = - (T1 + T2)
There are (1-t1) & sup2; + (1-t1 & sup2;) & sup2; = (1-t2) & sup2; + (1-t2 & sup2;) & sup2;
(according to (1,1), the distance from this point to two points is equal)
It is reduced to: (T1 & sup2; + T2 & sup2;) (T1 + T2) - (T1 + T2) - 2 = 0
That is: - (T1 & sup2; + T2 & sup2;) K + K-2 = 0
t1²+t2²=1-2/k
Because 2 (T1 & sup2; + T2 & sup2;) > (T1 + T2) & sup2; = K & sup2;,
So there is 1-2 / k > k & sup2 / 2
When k > 0, K ＾ 3-2k + 4

It is known that there are two symmetric points on the parabola y ^ 2 = x with respect to the straight line y = K (x-1) + 1, and the value range of the real number k is obtained

On the symmetry of the line y = K (x-1) + 1
As long as there is an intersection point between a parabola and a line, then at least two points are symmetrical about the line
[K(X-1)+1]^2=X
K^2X^2+2KX(1-K)+(1-K)^2=X
Derivative of K
2KX^2+2X-4KX-2+2K=0
K(X2-2X+1)+X=0
K(X-1)^2+X=0
K=-X/(X-1)^2

Set a = {(x, y) | y = x ^ 2 + MX + 2, m belongs to R}, B = {(x, y) | y = x + 1} if a intersects B = empty set, find the value range of real number M

A is an empty set
△1=m^2-8

Given the set a = {(x, y) / y = x ^ 2 + MX + 2} and B = {(x, y) / y = x + 1,0 ≤ x ≤ 2}, if a ∩ B ≠ empty set, find the value range of real number M
Y = x ^ 2 + MX + 2 and y = x + 1, 0 ≤ x ≤ 2
Simulink: x ^ 2 + (m-1) x + 1 = 0
So we can get that x ≠ 0, and M = - X-1 / x + 1,
When a ∩ B ≠ an empty set, X belongs to (0,2], - X-1 / X is less than or equal to - 2
So - X-1 / x + 1 is less than or equal to - 1, that is, M is less than or equal to - 1
M = - X-1 / x + 1,
When a ∩ B ≠ an empty set, X belongs to (0,2], - X-1 / X is less than or equal to - 2
thank you

When a ∩ B ≠ an empty set, B must not be an empty set, so x belongs to (0,2) (otherwise B is an empty set),
So x is a positive number, so - X-1 / x = - (x + 1 / x) 0, b > 0
Then a + b > = 2 * sqrt (a * b) a plus B is greater than or equal to 2 times of AB under the root sign

Given a = {x ∈ R | x2 + 2x + P = 0} and a ∩ {x ∈ R | x ＞ 0} = ∞, the value range of real number P is obtained

∩ {x ∈ R | x ＞ 0} = ∈, (1) if a = ∈, then △ = 4-4p ＜ 0, then p ＞ 1; (2) if a ≠, then a = {x | x ≤ 0}, that is, the roots of equation x2 + 2x + P = 0 are less than or equal to 0. Let two be X1 and X2, then △ = 4 − 4P ≥ 0x1 + x2 = − 2 ≤ 0x1x2 = P ≥ 0

Let a = {x | x 2 + (P + 2) x + 1 = 0, X ∈ r}, if a ∩ (0, + ∞) = ∞, find the value range of real number P

∩ (0, + ∞) = ∈, the equation x2 + (P + 2) x + 1 = 0 has no positive roots. There are two cases: (1) a = ∈, the equation has no solution, the discriminant is less than 0 (P + 2) 2-4 ＜ 0p2 + 4P ＜ 0-4 ＜ P ＜ 0 (2) a ≠ ∈ equation has two negative roots, the discriminant is greater than or equal to 0 and two and less than 0 (P + 2) 2-4 ≥ 0 and - (P + 2)

Known set a = {X / (1 / 2) ^ x ^ 2-x-6

Is this a = {X / (1 / 2) * X & # 178; - X-6

Given the set (x | x2 + 2 (a + 1) x + A2-1 = 0), B = (x | x2 + 4x = 0), a intersection B = a, find the value range of real number a

A={x|x^2+2(a+1)x+a^2-1=0}
B={x|x^2+4x=0}={0,-4}
A ∩ B = a means that a is a subset of B
And Δ = 4 (a + 1) ^ 2-4 (a ^ 2-1) = 8A + 8
①Δ

[a + 1 + (2-7a) divided by (a + 2)] divided by (the square of A-4 divided by the square of a + 4A + 4)

[a + 1 + (2-7a) / (a + 2)] divided by [(A & sup2; - 4) / (A & sup2; + 4A + 4)] = [(a + 1) (a + 2) + 2-7a] / (a + 2) divided by [(A-2) (a + 2) / (a + 2) & sup2;] = [(A & sup2; + 3A + 2 + 2-7a) / (a + 2)] = [(A-2) / (a + 2)] = (A & sup2; - 4A + 4) divided by (A-2) = (A-2) & sup2; divided by (A-2) = A-2