If we know the image crossing point (root 3 / 3, - 1 / 4) of the inverse function of the function y = a to the x power (a is greater than 0 and a is not equal to 1), then the value of a is

If we know the image crossing point (root 3 / 3, - 1 / 4) of the inverse function of the function y = a to the x power (a is greater than 0 and a is not equal to 1), then the value of a is

Nine
Because the inverse function is over (radical 3 / 3, - 1 / 4)
So root 3 / 3 = a ^ (- 1 / 4)
So a ^ - 1 = (root 3 / 3) ^ 4
So a = 9

Let f (x) = clnx + 1 And x = 1 is the extreme point of F (x) (1) If x = 1 is the maximum point of F (x), find the monotone interval of the function (expressed by C); (2) If f (x) = 0 has exactly two solutions, find the value range of real number C

(1) If we take the derivative function, we can get f '(x) = x2 + BX + C
X
∵ x = L is the maximum point of F (x), ᙽ f ′ (1) = 0
∴f′(x)＝(x−1)(x−c)
x，c＞1，b+c+1=0
When 0 ＜ x ＜ 1, f ′ (x) ＞ 0; when 1 ＜ x ＜ C, f ′ (x) ＜ 0; when x ＞ C, f ′ (x) ＞ 0;
The increasing interval of F (x) is (0,1), (C, + ∞); the decreasing interval is (1, c)
(2) (1) if C < 0, then f (x) decreases on (0, 1) and increases on (1, + ∞). If f (x) = 0 has exactly two solutions, then f (1) < 0,
∴1
2+b＜0，∴−1
2＜c＜0
② If 0 < C < 1, then f max (x) = f (c) = clnc + 1
2c2 + BC, f min (x) = f (1) = 1
2+b
∵ B = - 1-C, ᙽ Fmax (x) = f (c) = clnc + 1
2c2 + C (− 1 − C) < 0, fminimum (x) = f (1) = - 1
2-C, so that f (x) = 0 has only one solution;
③ If C > 1, then f min (x) = f (c) = clnc + 1
2c2 + C (− 1 − C) < 0, Fmax (x) = f (1) = - 1
2-C, so that f (x) = 0 has only one solution;
In conclusion, when f (x) = 0 has exactly two solutions, the value range of real number C is − 1
2＜c＜0

Let (1, b) of the graph (1 + 1) be equal to the inverse point (1,2) of the graph=

solution
F (x) = loga (BX + 1) image passes (1,1),
Because its inverse function is over (2,4), so f (x) image passes (4,2)
So we can get the following results:
loga (b+1)=1 (1)
loga (4b+1)=2 (2)
therefore
a=b+1
a^2=4b+1
The solution a = 3, B = 2 (a = 1, round off)
So a + B = 5

If the inverse function of the function f (x) = x + A / BX + C is 3x + 1 / 2x-1, then the values of a, B, C are,

According to the definition of inverse function, we can find that the inverse function of the original function is: F (x) = a-cx / BX-1 = 3x + 1 / 2x-1, and then we can know that: a = 1, B = 2, C = - 3 from the corresponding coefficient equality

Is the inverse function of the function f (x) = the sum of 1 + A ^ 2x under the radical sign - A ^ x [a > 0 and a is not equal to 1]?

a^x=t
y=(1+t^2)^0.5-t
t=(1-y^2)/2y
F-1 (x) = log (lower a) [(1-x ^ 2) / 2x]

Given the function f (x) = ax + K (a > 0 and a is not equal to 1) image crossing point (- 1,1), its inverse function F-1 (x) image crossing point (9,1), find the solution set of F-1 (x) < 0 Note: is a to the power of X + K; followed by F to the power of - 1 (x)

f(x)=ax+k
Bring in (- 1,1), (1,9) and find f (x),
Finding f (x) naturally leads to F-1 (x)
Finding F-1 (x) naturally leads to the solution set whose value is less than 0

If the function y = f (x) is the inverse function of the function y = ax (a > 0, and a ≠ 1), its image passes through a point（ a. A), then f (x) = () A. log2x B. log1 2x C. 1 2x D. x2

∵y=ax
⇒x=logay，
∴f（x）=logax，
∴a=loga
A=1
Two
⇒f（x）=log1
2x．
Therefore, B

The known function f (x) = a − x If the symmetry center of the inverse function image of X − a − 1 is (- 1,3), then the value of real number a is () A. 2 B. 3 C. -3 D. -4

The function f (x) = a − x
The symmetry center of the inverse function image of X − a − 1 is (- 1,3), so the symmetry center of the original function is (3, - 1),
The function is f (x) = a − X
x−a−1＝−1+−1
X − a − 1, so a + 1 = 3, so a = 2
Therefore, a

The image of the inverse function of the function f (x) = 2A ^ (x + 1) + 3 (a > 3), a is not equal to 1) must pass the fixed point? a> Wrong number 0

f(x)=2a^(x+1)+3
Constant over (- 1,5)
The inverse function is constant over (5, - 1)

(Shanghai Volume 8) for any positive number a not equal to 1, the image of the inverse function of function f (x) = loga (x + 3) passes through point P, then the coordinate of point P is______

The function f (x) = logax is constant over (1,0),
After the function f (x) = logax is shifted to the left by 3 units, the image of F (x) = loga (x + 3) is obtained
Therefore, the image of F (x) = loga (x + 3) crosses the fixed point (- 2,0),
In addition, the image of two functions which are inverse functions of each other is symmetric with respect to the straight line y = X,
So the image of its inverse function crosses the fixed point (0, - 2)
The answer is: 2