How to find f (x) when a mathematical function in grade one of senior high school has f (x-1 / x + 1) = - X-1

How to find f (x) when a mathematical function in grade one of senior high school has f (x-1 / x + 1) = - X-1

Let a = (x-1) / (x + 1)
ax+a=x-1
x-ax=a+1
x=(1+a)/(1-a)
So f (a) = - (1 + a) / (1-A) - 1 = - 2 / (1-A)
So f (x) = - 2 / (1-x)
Given that the function f (x) satisfies f (loga x) = (x-x ^ - 1) / (a ^ 2-1), where a > 0 and a is not equal to 1. Find the analytic expression of F (x)
Let y = loga x, then x = a ^ y
f(y)=(a^y-a^-y)/(a^2-1)
f(x)=(a^x-a^-x)/(a^2-1)
F (x + 2) = 1 / F (x), f (1) = 5; ask f (f (x))?
Let x = 1, get F3 = 1 / 5, x = 3, F5 = 5, you can draw an image, you can take y = asin (AX + b) and then solve the problem
The function y = f (x) (x is not equal to 0) is an odd function and an increasing function when x belongs to (0, + OO). If f (1) = 0, find the inequality f [x (x-1 / 2)]
The idea is absolutely right
But the solution set of inequality 2 is wrong
It should be solved in detail by matching method, as shown in the figure
If f (SiNx) = sin3x, then f (cosx) =?
A.-cos3x
B.cos3x
C.sin3x
D.-sin3x
f(cosx)=f[sin(π/2-x)]
=sin[3(π/2-x)]
=sin(3π/2-3x)
=sin[π+(π/2-3x)]
=-sin(π/2-3x)
=-cos3x
Why do you come to this question again
Let SiNx = u, then (cosx) ^ 2 = (1-u ^ 2),
f(u)=sin3x=sin(2x+x)=sin2xcosx+cos2xsinx=2sinx(cosx)^2+(1-2(sinx)^2)sinx=2u(1-u^2)+(1-2u^2)u=3u-4u^3,
Then f (cosx) = 3cosx-4 (cosx) ^ 3
3cosx-4*(cosx)^3
A
f(cosx)
=f[sin(π/2-x)]
=sin[3(π/2-x)]
=sin(3π/2-3x)
=-cos3x
Given that f (x) is an increasing function on the definition (0, positive infinity), and f (x / y) = f (x) - f (y), f (2) = 1, the solution of the inequality f (x) - f (1 / (x-3)) is less than or equal to 2
Because the field of F (x) is (0, positive infinity), so
1 / (x-3) > 0, x > 3
F (4) - f (2) = f (4 / 2) = f (2) = 1, so f (4) = 2
From F (x) - f (1 / (x-3))
F (2) = f (2 / 1) = f (2) - f (1) = 1, so f (1) = 0
And f (x) - f (1 / (x-3))
=f[x(x-3)]
F (SiNx) = sin3x, f (cosx) = 3cosx-4 (cosx) ^ 3 or = - cos3x
Both methods seem to be right. Why is the other wrong?
I don't know how you do it, unless you shoot it with your mobile phone;
My approach is:
f(cosx)=f[sin(π/2-x)]=sin3(π/2-x)=sin(3π/2-3x)= - cos3x
That's right,
,f(cosx)=3cosx-4cos³x
This formula is made by using f (cosx) = cos3x
Only f (SiNx) = sin3x,
So when we meet f (cosx), we have to dress up as f (COS.) to have output,
If the odd function y = f (x) (x ≠ 0), when x ∈ (0, + ∞), f (x) = X-1, then the solution of inequality f (x-1) < 0 is______ .
∵ function y = f (x) is an odd function. When x > 0, f (x) = X-1, when x < 0, - x > 0, f (- x) = - f (x) = - X-1, i.e. f (x) = x + 1, when X-1 > 0, i.e. x > 1, f (x-1) = X-2, the original inequality is transformed into X-2 < 0, and the solution set of the original inequality is (1,2). When X-1 < 0, i.e. x < 1, f (x-1) = x, and f (1-x) = - x is an odd function In this case, the solution set of the original inequality is (- ∞, 0) ∪ (1,2). So the answer is: (- ∞, 0) ∪ (1,2)
Sin3x / sinx-cos3x / cosx reduction
Original formula = (sin3x * cosx-cos3x * SiNx) / (SiNx * cosx)
= 2sin(3x-x)/sin2x=2
sin3x/sinx-cos3x/cosx=-(cos3x/cosx-sin3x/sinx)
=-cos(3x-x)
=-Cos2x: you are wrong
If f (x) = a ^ (x + b), (a > O, a is not equal to 1) satisfies f (x + y) = f (x) f (y), and f (3) = 8, find f (x)
Let x = y = 1
f(x+y)=f(2)=a^(b+2)=f(x)*f(y)=a^(b+1+b+1)=>b=0
F (3) = a ^ 3 = 8 (a > O, a is not equal to 1)
A=2
f(x)=2^x