Given that the function f (x) = ax + B / 1-x2 is an odd function defined on (- 1,1), and f (1 / 2) = 3 / 4, the analytic expression of function f (x) is obtained

Given that the function f (x) = ax + B / 1-x2 is an odd function defined on (- 1,1), and f (1 / 2) = 3 / 4, the analytic expression of function f (x) is obtained

The function f (x) = (AX + b) / (1-x ^ 2) is an odd function defined on (- 1,1)
So f (0) = b = 0
Because f (1 / 2) = (A / 2) / (3 / 4) = 4 / 3
So a = 2
So f (x) = (2x) / (1-x ^ 2)

If f (x) satisfies f (x 2) = x2 3, then f (x)=
f(x+2)=x2+3

Let t = x + 2
x = t-2
f(x+2) = f(t) = (t-2)2 +3
So f (x) = (X-2) 2 + 3

Given the function f (x) = x2 + 2ax-3, if f (a + 1) - f (a) = 9, find the value of A
Process, 20 points

f(x)=x²+2ax-3
f(a+1)=(a+1)²+2a(a+1)-3=a²+2a+1+2a²+2a-3=3a²+4a-2
f(a)=a²+2a²-3=3a²-3
f(a+1)-f(a)=(3a²+4a-2)-(3a²-3)=4a+1=9
The solution is a = 2

If we know that f (x) = 4log, the logarithm of 3 with 2 as the base + 233, then the value of F (2) + F (4) + F (8) +. + F (2 to the power of 8)

f(3^x) = 4x(log3/log2) + 233 f(3^x) = 4log(3^x)/log2 + 233
So f (x) = 4 logx / log2 + 233
So f (2) + F (4) +... + F (256) = 4 (1 + 2 + 3 +... + 8) + 233 * 8 = 2008

It is known that f (3 to the power of x) is equal to 4xlog2 (3) plus 233. Then what is f (2) plus f (4) plus ellipsis to the power of 8 of F (2)
Like the title,

Key steps: let the x power of 3 = 2, x = log23, 2 up and 3 down
Because I don't know what your 4xlog2 (3) means

It is known that the domain of the function y = (log2x) ^ 2-2log2x + 3 is [1,4]
Finding the maximum and minimum of a function

1≤x≤4
log2 1≤log2 x≤log2 4
0≤log2 x≤2
Let t = log2 x, t ∈ [0,2], then
y=t²-2t+3=(t-1)²+2
When t = 1, y is the minimum, y (min) = 2
When t = 0 or T = 2, y is the maximum, y (max) = 3

Let f (x) = 2log2x-log2 (x-1), then the domain of F (x) is_ The minimum value of F (x) is__ .

The definition of the definition of the definition of the definition of X \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\923; 2 Λ f (x)

Given the function f (x) = 3-2log2x, G (x) = log2x 1
Given the function f (x) = 3-2log2x, G (x) = log2x 1. When x ∈ [1,4], find the range of function H (x) = [f (x) + 1] × g (x). 2. If for any x ∈ [1.4], the inequality f (X & # 178;) × f (x under the root sign) > k × g (x) holds, find the range of real number K

Let t = log2x, t belongs to [0,2]
We get H (T) = (4-2t) t = - 2T ^ 2 + 4T
When t = 1, there is a maximum of 2; when t = 0 or 2, there is a minimum of 0
So the range of H (x) is [0,2]
(2) The inequality is transformed into (3-4t) (3-T) > KT
When t = 0, 9 > 0 holds
When 0

The minimum value of the maximum value m (a) of the function f (x) = | x2-a | in the interval [- 1,1] is ()
A. 14B. 12C. 1D. 2

The function f (x) is an even function, so it is necessary to discuss the range of M (a) only in the range of X ∈ [0,1]; & nbsp; & nbsp; 1 ＞ when 0 ＜ a ＜ 1, then when a ≤ 0, the function f (x) monotonically increases in [0, 1], m (a) = f (1) = | 1-A | = 1-A ≥ 1. When a ＞ 0, the function f (x) monotonically decreases in [0, a] and monotonically increases in [a, 1], so the maximum value of F (x) in [0, a] is f (0) = a, and the maximum value of F (x) in [a, 1] is f (1) = 1-A, from F (1) ＞ f (0), i.e. 0 ＜ a ＜ 12 When a ∈ (0,12), m (a) = f (1) = 1-A. similarly, when a ∈ [12,1], m (a) = f (0) = A. when a ≥ 1, the function is a decreasing function in [0,1], so m (a) = f (0) = A & nbsp; when a ≤ 0, f (x) = | x2-a | = x2-a, and is an increasing function in [0,1], so m (a) = f (1) = 1-A. to sum up, m (a) = 1-A, a < 12; & nbsp; & nbsp; M (a) = a, a ≥ 12, so m (a) is a decreasing function in [0,12] and an increasing function in [12,1]. In conclusion, the minimum value of M (a) is m (12) = 12, so B is selected

Find the maximum value of the function y = 1 / 2 + SiNx + cosx

According to the sum difference product formula sin (a + b) = sinacosb + sinbcosay = 1 / 2 + SiNx + cosx = 1 / 2 + radical 2 (radical 2 / 2 * SiNx + radical 2 / 2 * cosx) = 1 / 2 + radical 2 [cos (π / 4) * SiNx + sin (π / 4) * cosx] = 1 / 2 + radical 2Sin (x + π / 4) because sin (x + π / 4) ≤ 1y, the maximum value = 1 / 2 + radical 2