1. If f (2x) = 3x-1 and f (a) = 5, then a =

1. If f (2x) = 3x-1 and f (a) = 5, then a =

2x=a...（1）
3x-1=5.（2）
3x=6
X=2
Substituting (1) a = 2 * 2 = 4
Let a = 2x
Then f (2x) = f (a) = 3x-1 = 5
X=2
A=4
Let 2x = a, then x = A / 2 and f (2x) = 3x-1
f（a）=3a/2-1=5
The solution is a = 4
If 2x = y, then x = Y / 2
f(Y)=3*y/2-1
=1.5y-1
I.e. f (x) = 1.5x-1
f(a)=1.5a-1=5
A = 4
It is known that the domain of definition of function f (x) is r, and for any real number x, it satisfies: F (2 + x) = f (2-x), f (7 + x) = f (7-x), if f (5) = 9,
Find the value of F (- 5)
The values of 2 + X and 2-x are equal, which indicates that FX is symmetric with respect to 2
Similarly, about 7 symmetry
On the symmetry of 2 and 7, it shows that the function is a periodic function with period of 5
f（-5）=f(0)=f(5)=9
f(-5)=f(2-7)=f(2+7)=f(9)=f(7-2)=f(5)=9
f(-5)=f(2-7)=f(2+7)=f(7-2)=F(5)=9
Given f (x) = 6x ^ 5 + 4x ^ 4-3x ^ 3 + 2x ^ 2, G (x) = f (x) / (- 1 / 2x)
Find the expression of (1). G (x)
(2) . g (1) + G (- 1)
solution
g(x)=f(x)/(-1/2x)
=-2f(x)/x
=-2(6x^4+4x³-3x²+2x)
=-12x^4-8x³+6x²-4x
g(1)=-12-8+6-4=-18
g(-1)=-12+8+6+4=6
g(1)+g(-1)
=-18+6
=-12
Let f (x) be defined as R. when x > 0, f (x)
In fact, you have to associate exponential function with this kind of function
(1) Let x = y = 0
∴f(0)=f(0)*f(0)
∵f(0)≠0
∴f(0)=1
When x0,
∴f(-x)＜1
∵1=f(0)=f(x+(-x))=f(x)f(-x)
f(x)=1/f(-x)＞1
∴f(x)＞1＞0
Let X1 < x2
f(x2)-f(x1)=f[(x2-x1)+x1]-f(x1)
=f(x2-x1)*f(x1)-f(x1)
=f(x1)*[f(x2-x1)-1]
∵x2-x1＞0
∴f(x2-x1)＜1
∴f(x2-x1)-1＜0
And ∵ f (x1) > 0
∴f(x2)-f(x1)＜0
∴f(x2)＜f(x1)
Ψ f (x) is a decreasing function
(2) For f (x) f (3x-1), the problem can be reduced to f (x + 3x-1)
f(2)=f(1+1)=f(1)*f(1)=1/9
∴f(1)=1/3
f(1)*f(2)=1/27=f(3)
The inequality can be reduced to f (x + 3x-1) < f (3)
∵ f (x) is a decreasing function
∴x+3x-1＞3
The solution is x > 1
Do not understand welcome to ask
Because f (x + y) = f (x) f (y)
So f (x) f (3x-1) = f [(x) + (3x-1)] = f (4x-1)
Let x = 1, y = 1, then f (2) = [f (1)] ² = 1 / 9
Because let x = y = 1 / 2, we can get f (1) = [f (1 / 2)], # 178;
So f (1) = 1 / 3 negative rounding off
So f (3) = f (1 + 2) = f (1) f (2) = f (1) f (1 + 1) = f (1) f (1) = 1 / 27
The original inequality can be reduced to f (4x-1) 3, that is, x > 1
If f (x) = 3x ^ 2-2x + 1, G (x) = 3x-2. Find f (g (x)) =? And G (f (x)) =?
If f (x) = 3x ^ 2-2x + 1, G (x) = 3x-2
F (g (x)) =? And G (f (x)) =?
f(g(x)) = f(3x-2) = 3(3x-2)^2 - 2(3x-2) + 1 = 27x^2 - 42x +17
g(f(x)) = g(3x^2 -2x+1) = 9x^2 - 6x +1
For any real number T, f (5 + T) = f (5-T). Compare f (- 1) f (9) f (13)
F (9) = f (5 + 4) = f (5-4) = f (1), and 1 is in the given interval; f (13) = f (5 + 8) = f (5-8) = f (- 3), then - 3 is in the given interval
Given that f (x) = 3x-1, G (x) = 2x + 3, and f [H (x)] = g (x), then H (x)=___
f(x)=3x-1,g(x)=2x+3 f[h(x)]=g（x）
F [H (x)] = 3H (x) - 1 = g (x) = 2x + 3, i.e
3h(x)-1=2x+3
We get H (x) = (2x + 4) / 3
（2x+4）/3
Solution: H (x) = 2 / 3x + 4 / 3. Because f [H (x)] = g (x). [substituting H (x) as x into f (x)], so 3H (x) - 1 = 2x + 3. So h (x) = (2x + 3 + 1) / 3 = 2 / 3x + 4 / 3
It is known that the function f (x) is an odd function with the domain R, and its image is symmetric with respect to the straight line x = 1. (1) find the value of F (0). (2) prove that the function f (x) is a periodic function
(1) Because the function f (x) is an odd function with the domain R, f (- x) = - f (x), when x = 0, f (- 0) = - f (0), so f (0) = 0. (2) because the function is symmetric with respect to x = 1, f (1 + x) = f (1-x), that is, f (1 + x) = f (1-x) = - f (x-1), so f (x + 2) = - f (x), that is, f (x + 4) = f (x). Therefore, the function is a periodic function with a period of 4
F (x) = 2 ^ (- x ^ 2 + 3x + 2) range
-x²+3x+2
=-(x-3/2)²+17/4≤17/4
So 0
Given the function y = f (x) of the domain of definition on the set of real numbers, f (x + y) = f (x) + F (y) holds for any x, y, and f (x) < 0 holds when x > 0
f(x)=-x
So the range on [M, n] is [- N, - M]
It is easy to get that f (x) = x * f (1), so F1 = - 1, and f (n + 1) - f (n) = F1 is less than 0, so the function is a decreasing function on Z, so the range is [- N, - M]
m=3215