What is the maximum value of the function FX = x ^ 2-2ax + A in the interval x ＜ 1 and X ＞ 1

What is the maximum value of the function FX = x ^ 2-2ax + A in the interval x ＜ 1 and X ＞ 1

F (x) = (x-a) ^ 2-A ^ 2 + A, where x

Given the function FX = (2ax-1) / (2x + 1), when a = 1, find the monotone interval of FX

When a = 1 is introduced into the function, it is transformed into FX = (2x-1) / (2x + 1)
Where x is not equal to - 1 / 2, otherwise it has no real meaning
f ’x=[（2x-1）' (2x+1)-(2x+1）' (2x-1)]/(2x+1)^2
f ’x=[2(2x+1)-2(2x-1)]/(2x+1)^2
f ’x=4/(2x+1)^2>0
So the monotone interval is (- infinity, - 1 / 2), (- 1 / 2, + infinity)

Please help to solve the following mathematical problems: let an original function of function f (x) be xlnx: (1) ∫ XF (x) DX

f(x)=(xlnx)'=1+lnx
∫xf(x)dx=∫x(1+lnx)dx
=∫xdx+∫xlnxdx
=x^2/2+∫lnxd(x^2/2)+C
=x^2/2+lnx*x^2/2-∫x/2 dx+C
=1/4*x^2+1/2*x^2lnx+C
Please open a new question,

A function problem of senior one: given f (x) + 2F (1 / x) = 2x + 1, find the analytic expression of F (x)
It's better to have a process or solution and a way to solve such problems

X is replaced by 1 / X
f(x)+2f(1/x)=2x+1
f(1/x)+2f(x)=2/x+1
Simultaneous equations, eliminate f (1 / x), you can get the answer
f(x)=4/(3x) -2/3*x + 1/3
I don't understand

Elimination method of higher one function if 3f (x-1) + 2F (1-x) = 2x to find f (x)
Let t = X-1, then the original formula of x = t + 1 becomes 3f (T) + 2F (- t) = 2 (T + 1) and the original formula of - t becomes t becomes 3f (T) + 2F (T) = 2 (1-T). By eliminating f (- t) above, we can get f (T) = 5 of 2T + 2, so f (x) = 5 of 2x + 2
Why eliminate f (- t) to get 2x + 2 / 5, know how to shout, thank you ~ ~ ~ ~ detailed conversion process, the best multi text expression~~

Let x = X-1
Then 3f (x) + 2F (- x) = 2 (x-1) (1)
Where x = - x
Then 3f (- x) + 2F (x) = 2 (- x-1) (2)
(1) * 3 - (2) * 2
9f(x)-4f(x)=6(x-1)+4(x+1)
5f(x)=10x-2
f(x)=2x-2/5

A function problem of senior one: given f (x) = 2x + 3, G (2x-1) = f (the square of x minus 1), find the analytic formula of G (x)

G (2x-1) = f (x ^ 2-1) = 2 (x ^ 2-1) + 3 = 2x ^ 2 + 1, let t = 2x-1, so x = (T + 1) / 2, so g (2x-1) = g (T) = 2 [(T + 1) / 2] ^ 2 + 1 = (1 / 2) * T ^ 2 + T + 3 / 2, so g (x) = (1 / 2) * x ^ 2 + X + 3 / 2. Method 2: let t = 2x-1, so x = (T + 1) / 2g (2x-1) = g (T) = f [(T ^ 2 + 2T + 1) / 4-1] = f [(T + 1) / 4-1]

Given that the function f (x) satisfies 2F (x) - f (- x) = - x ^ 2 + 4x, try to find the expression of F (x)

2F (- x) - f (x) = - X & # 178; - 4x 4f (x) - 2F (- x) = - 2x & # 178; + 8x is added to get f (x) = - X & # 178; + 4x / 3

Given that the function f (x) is a linear function and satisfies the relation 3f (1 + x) - 2F (1-x) = 4x + 3, what methods do you have to find out the expression of F (x)
We know one (let y = ax + b)

∵3f(1+x)-2f(1-x)=4x+3
∴3f(1-x)-2f(1+x)=-4x+3
Two equations deduce that f (x + 1) = 4 / 5x + 3 = 4 / 5 (x + 1) + 11 / 5
∴f(x)=4/5x+11/5

Let's use the descriptive method to represent the following sets! The set of all real numbers less than 3 and greater than 5. The set of all real numbers of equation x2-5x + 6 = 0

The set {x | X5} consists of all real numbers less than 3 and greater than 5; the set {x | x2-5x + 6 = 0} consists of all real numbers of equation x2-5x + 6 = 0;

Given that f (x) satisfies f (4x-1) = 2F (x) + 6x + 1, the expression of F (x) is obtained;
If G (x) = {- x2-4x-4, X

Let f (x) = ax + B
Because f (4x-1) = 2F (x) + 6x + 1,
therefore
a(4x-1)+b=2ax+2b+6x+1
(2a-6)x-a-b-1=0
X holds for everything, so
2a-6=0
-a-b-1=0
thus
a=3,b=-4
f(x)=3x-4.