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Given that the inverse function g (x) of the function f (x) = 2x satisfies g (a) g (b) = 4, find the minimum value of 1 part of a and 1 part of B
The inverse function g (x) = x / 2 of the function f (x) = 2x
If G (a) g (b) = 4, then AB / 4 = 4, ab = 16
1 / A + 1 / b > = 2 roots, 1 / AB = 2 * 1 / 4 = 1 / 2, that is, the minimum value is 1 / 2
Let F-1 (x) be the inverse function of the function f (x) = log2 (x + 1). If [1 + F-1 (a)] [1 + F-1 (b)] = 8, then the value of F (a + b) is the value of F (a + b) Urgent! Online, etc The - 1 in the title represents the inverse function 2 in the lower right corner of the log
The inverse function is F-1 (x) = 2 ^ x + 1
The substitution formula is (2 ^ a) (2 ^ b) = 2 ^ (a + b) = 8
So a + B = 3
f(a+b)=log2 (4)=2
The known function f (x) = log2 (x ^ 2 + 1) (x
Negative radical 3
Because 2 ^ y = x ^ 2 + 1, X
Function f (x) = log2 (x + 1), (x > - 1), find its inverse function
y=log2(x+1)
2^y=x+1
x=2^y-1
The inverse function is y = 2 ^ X-1
Inverse function of function f (x) = log2 (2 ^ x-1)
To find the inverse function of a function, the first thing you need to know is the relationship between a function and its inverse function. Their relationship is very clear, that is, the definition domain of a function is the value range of another function, and the value range of this function is the definition domain of another function
Given that the function f (x) = log2 (x), X belongs to [2,8]. The minimum value of the function g (x) = f (x) ^ 2-2af (x) + 3 is h (a) (1). Find H (a)
f(x)∈[1,3]
g(x)=[f(x)-a]^2+3-a^2
This is not a quadratic function to find the minimum value, the definition domain is [1,3], the interval is invariant, the symmetry axis changes, and then discuss where the symmetry axis is (write the point is this: t ∈ [1,3], G (T) = [T-A] ^ 2 + 3-A ^ 2)
1.a≤1,h(a)=(g(1)-a)^2+3-a^2
2.1＜a＜3,h(a)=3-a^2
3.a≥3,h(a)=(g(3)-a)^2+3-a^2
Given √ 2 ≤ x ≤ 8, find the maximum and minimum values of the function f (x) = (log2x / 2). (log24 / x)
Because √ 2 ≤ x ≤ 8
Therefore: 1 / 2 ≤ log2 ^ x ≤ 3, let t = log2 ^ X
Therefore: 1 / 2 ≤ t ≤ 3,
Also: F (x) = (log2 ^ X / 2). (log2 ^ 4 / x)
=( log2^ x- log2^ 2)( log2^ 4- log2^ x)
=( log2^ x- 1)( 2- log2^ x)
=(t-1)(2-t)
=-t²+3t-2
=-(t-3/2)²+1/4
Therefore: when t = 3 / 2, take the maximum value of 1 / 4, at this time log2 ^ x = 3 / 2, x = 2 √ 2
When t = 3, take the minimum value - 2, and log2 ^ x = 3, x = 8
The function f (x) = 2 ^ 2 + a · log2 (x ^ - 2) + B has a minimum value of 1 when x = 1 / 2
The function f (x) = 2 [log2 (x)] ^ 2 + a · log2 [x ^ (- 2)] + B has a minimum value of 1 when x = 1 / 2
f'(x)=4*[log2(x)][1/(x ln2)]+a·[-2x^(-3)]/[x^(-2)*ln2]=
=4*[log2(x)]/(x ln2)-2ax^(-3)/[x^(-2)*ln2],
Function has a minimum value, it must be the minimum value first
f'(x)=4*[log2(x)]/(x ln2)-2ax^(-3)/[x^(-2)*ln2]=0,
2[log2(x)]/(x ln2)=ax^(-3)/[x^(-2)*ln2],
2[log2(x)][x^(-2)*ln2]=ax^(-3)(x ln2),
Let this minimum be the minimum, where x = 1 / 2, then x = 1 / 2: x = 1 / 2
2[log2(1/2)][(1/2)^(-2)*ln2]=a(1/2)^(-3)((1/2) ln2),
-2[4ln2]=a*4ln2,
a=-2.
The function f (x) = 2 [log2 (x)] ^ 2 + a · log2 [x ^ (- 2)] + B has a minimum value of 1 when x = 1 / 2,
Then f (1 / 2) = 2 [log2 (1 / 2)] ^ 2 + a · log2 [(1 / 2) ^ (- 2)] + B = 1,
A = - 2,
2-2*2+b=1,
b=3.
Let f (x) = log2 (| X-1 | + | X-5 - a) (1) When a = 2, find the minimum value of function f (x); (2) When the definition domain of function f (x) is r, find the value range of real number a
The definition domain of function satisfies | X-1 | + | X-5 | a | 0, that is | X-1 | + | X-5 | a,
(1) When a = 2, f (x) = log2 (| X-1 | + | X-5 - 2)
Let g (x) = | X-1 | + | X-5 |, then G (x) = | x − 1 | + | x − 5 |
2x−6(x≥5)
4(1＜x＜5)
6 − 2x (x ≤ 1). (3 points)
G (x) min = 4, f (x) min = log2 (4-2) = 1. (5 points)
(2) According to (I), the minimum value of G (x) = | X-1 | + | X-5 | was 4, 7 points | X-1 | + | X-5 | a ﹤ 0,
∴a＜4
The value range of a is (- ∞, 4). (10 points)
If the inverse function of function f (x) is f ^ (- 1) (x) = log2 (x), then f (x)=
The original function is y = 2 ^ X
It's an X power of 2
Either way, definition or drawing
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