Given the odd function f (x) defined on R, if x > 0, f (x) = (1 / 2) ^ x, find the analytic expression of F (x)

Given the odd function f (x) defined on R, if x > 0, f (x) = (1 / 2) ^ x, find the analytic expression of F (x)

Odd function f (x) = - f (- x)
F (0) = - f (0) so f (0) = 0
For x0, = 0,

F (x) = log (a) (1 + x), the function value on [1, + ∑) always has | f (x) | > 2, and the value range of a is obtained

Classification discussion:
① 02
It can be found that when 0 1, | f (x) | = log (a) (1 + x) > 2
We can get the root of a > 2

For X ∈ R, the function f (x) = x ^ 2-4ax + 2A + 30 (a ∈ R) is a non negative number, and the range of the roots of the equation x / (a + 3) = | A-1 | + 1 is obtained
Let f (x) = x ^ 2-4ax + 2A + 30 (a ∈ R) be nonnegative,
From the discriminant > = 0, we get - 5 / 2

Discussion on Classification:
-5/2

Let the values of all real numbers x, y = x ^ 2-4ax + 2A + 6 be nonnegative, and find the maximum value of function f (a) = 2-A * | a + 3 |

Y = x ^ 2-4ax + 2A + 6 is a parabola with the opening upward
Its value is always nonnegative, so the minimum value is nonnegative
y = x^2 - 2 * 2a * x + (2a)^2 - (2a)^2 + 2a + 6
= (x - 2a)^2 - 2(2a^2 -a -3)
The minimum value is - 2 (2a ^ 2 - A - 3)
-2 (2a^2 - a -3) ≥ 0
(2a -3)(a + 1) ≤0
-1 ≤ a ≤ 3/2
In this range, a + 3 is always greater than 0
f(a) = 2 - a(a+3)
= -a^2 - 3a + 2
= - [a^2 + 3a -2]
= - [(a + 3/2)^2 - 9/4 - 2]
= 17/4 - (a + 3/2)^2
F (a) is a parabola with a = - 3 / 2 as its vertex
The interval - 1 ≤ a ≤ 3 / 2 is on the right side of a = - 3 / 2
The best value is
f(-1) = 2 - (-1)*|-1 + 3| = 4
The minimum value is
f(3/2) = 2 - (3/2)|3/2 + 3| = -19/4

For any real number x, the value of the function f (x) = x ^ 2-4ax + 2A + 30 is non negative. The value range of the equation x / (a + 3) = | A-1 | + 1 about X is obtained

The value of function f (x) = x ^ 2-4ax + 2A + 30 is non negative, that is, the discriminant value of equation x ^ 2-4ax + 2A + 30 = 0 is not less than 0
（-4a）^2-4（2a+30）

The maximum value of the function y = log (a) x (2 ≤ x ≤ 5) is greater than the minimum value by 1, and the value of the base a is obtained
Note: log (a) x represents the logarithm of X with a as the base
Pay attention to the format when solving the problem. Be sure to write it step by step, just like on the answer sheet,

If a > 1, the maximum value is when a = 5, and the minimum value is when a = 2, namely log (a) 5-log (a) 2 = 1. A = 5 / 2,
If a

If the difference between the maximum value and the minimum value of the function f (x) = loga (x + 1) on [0, 3] is 2, then the value of a is___ ．

If a > 1, f (x) = loga (x + 1) increases monotonically on [0,3], f (x) max = loga4 = 2loga2, f (x) min = loga1 = 0, ∵ f (x) max-f (x) min = 2, ∵ 2loga2-0 = 2, ∵ loga2 = 1, so a = 2; if 0 < a < 1, f (x) = loga (x + 1) decreases monotonically on [0,3], the same as

Find the maximum and minimum value of function f (x) = [Log1 / 4 base x index] bracket 2-log1 / 4 base x (2 power) index + 5 in the range of [2,4]

Let t = Log1 / 4 base x
Because x is a decreasing function based on Log1 / 4, so - 1

Given the function f (x) = log small 3 (4 / x + 2), then the value of X in the equation f ^ negative 1 (x) = 4 is? 1

The function f (x) = log small 3 (4 / x + 2),
3^f(x)=4/x+2
x=4/(3^f(x)-2)
The inverse function is
F ^ negative 1 (x) = 4 / (3 ^ X-2) = 4
3^x-2=1
x=1

Log 25 (2x) - log 5 (8) = 0, how much is x to solve the equation?

Log5 '(2 * 2x) = log5'8
SO 2 * 2x = 8
x=2
Hope to adopt!