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x>0 f(x)=(1/2)^x ,0

Given that the function f (x) = 1 + logax (a > 0 and a ≠ 1), F-1 (x) is the inverse function of F (x). If y = F-1 (x) + A is over (2,1), calculate the value of A A.3 B.2 C1/2 D1/3

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Given that the function y = logax has a point (2,3) on the image, then there must be a point () Please explain

If the function y = logax has a point (2,3) on the image, then there must be a point (3,2) on the image of its inverse function
Because the image of the inverse function is symmetrical to the image of the original function with respect to the line y = x, the symmetry point of (2,3) with respect to y = x is (3,2)

The function f (x) = logax 3 is a minus function It's logax + 3

Logax + 3 is a minus function. Does it mean the inverse function of 0
g(x)=a^(x-3)

If the function f (x * 2-3) = logax * 2-6-x * 2, find the inverse function F-1 (x) of the function f (x) 2 if f (x) > = loga2x, find the value range of X If F-1 (x)

f(x)=loga(x+3)/(3-x)
F-1 (x) = 3 (x of A-1) / (X-Power of a + 1)
2. Discussion
a>1,(x+3)/(3-x)>=2x
Zero
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It is known that the inverse function of the function f (x) = 2x is f − 1 (x). If f − 1 (a) + F − 1 (b) = 4, then 1 A+1 The minimum value of B is () A. 1 B. 1 Two C. 1 Three D. 1 Four

The inverse function of the function y = 2x is y = F-1 (x) = log2x,
So F-1 (a) + F-1 (b) = 4 is log2a + log2b = 4,
AB = 16 (a, b > 0)
One
A+1
b≥2
One
a×1
B=1
2, (if and only if a = b)
Therefore, B

Given the function f (x) = log3 (3 ^ x-1), if F-1 (x) is the inverse function of F (x), Let f (x) = F-1 (2x) - f (x), find the minimum value of function f (x)

(x) = log3 (3 ^ 2x + 1) / (3 ^ x-1)] let 3 ^ x = k = K (3 ^ 2x + 1) / (3 ^ x-1)] let 3 ^ x = k, then [(3 ^ 2x + 1) / (3 ^ x-1)] let 3 ^ x = k, then [(3 ^ 2x + 1) / (3 ^ x-1)] = (K + 1) / (k-1) = K + 1 + 1 (k-1 + 2) / (k-1) = K + 1 + 1 / (k-1) + (k-1) + 1 / (k-1) + 23 (k-1) + 1 / (k-1) + 23 (k-1) + 23 (k-1) + 23 (k-1) + 23 (k-1) + 23 (k-1) + 23^ x = k > 0k-1 > - 1 but log3

If f (x) = the image crossing point (- 1, n) of the inverse function of logmx, then the minimum value of 3N + m is () A. 2 Two B. 2 C. 2 Three D. 5 Two

From the image crossing point (- 1, n) of the inverse function of the function f (x) = logmx,
The graph crossing point (n, - 1) of the original function is logmn = - 1, ﹥ m > 0, n > 0, Mn = 1,
From the mean inequality, 3N + m ≥ 2 is obtained
3mn＝2
3, if and only if 3N = M,
Therefore, C

Let f (x) = (1 / 3) ^ x, its inverse function is y = g (x) 2. When x ∈ [- 1,1], find the minimum value H (a) 3 of the function y = [f (x)] ^ 2-2af (x) + 3 The inverse function of F (x) = (1 / 3) ^ x is y = g (x) 2. When x ∈ [- 1,1], find the minimum value H (a) of the function y = [f (x)] ^ 2-2af (x) + 3 3. Whether there is a real number m > n > 3, so that the definition field of function y = H (x) is [n, M] and the value field is [n ^ 2, m ^ 2], if there is, find the value of M, N, if not, explain the reason

Formula y = [f (x) - A] ^ 2-A ^ 2 + 3
When x ∈ [- 1,1], 1 / 3 ≤ f (x) ≤ 3
When a ≥ 3, the minimum H (a) = 12-6a is taken at f (x) = 3
When a ≤ 1 / 3, the minimum H (a) = 28 / 9-2a / 3 is taken at f (x) = 1 / 3
When 1 / 3 < a < 3, the minimum H (a) = 3-A ^ 2 is taken at f (x) = a
m> If n > 3, the definition domain of function y = H (x) is [n, M],
Then y = H (x) = 12-6x,
At this time, the function decreases monotonically
So h (n) = m ^ 2, H (m) = n ^ 2,
That is 12-6n = m ^ 2,
12-6m=n^2,
By subtracting the two formulas, 6 (m-n) = m ^ 2-N ^ 2,
Because m > N, 6 = m + n,
∵m>n>3,
M + n = 6 is impossible
Therefore, there is no real number m > n > 3

It is known that the inverse function of the function f (x) = 2x is f − 1 (x). If f − 1 (a) + F − 1 (b) = 4, then 1 A+1 The minimum value of B is () A. 1 B. 1 Two C. 1 Three D. 1 Four

The inverse function of the function y = 2x is y = F-1 (x) = log2x,
So F-1 (a) + F-1 (b) = 4 is log2a + log2b = 4,
AB = 16 (a, b > 0)
One
A+1
b≥2
One
a×1
B=1
2, (if and only if a = b)
Therefore, B