The function y = f (x) is an even function on (- ∞, + ∞). When x ≥ 0, f (x) = x ^ 2-2x-3. The expression of y = f (x) is written in the form of piecewise function There should be a process

The function y = f (x) is an even function on (- ∞, + ∞). When x ≥ 0, f (x) = x ^ 2-2x-3. The expression of y = f (x) is written in the form of piecewise function There should be a process

x0
So f (- x) is suitable for f (x) = x ^ 2-2x-3
So f (- x) = x ^ 2 + 2x-3
Even function f (x) = f (- x)
therefore
No more braces
f(x)=x^2-2x-3 x>=0
f(x)=x^2+2x-3 x

Given that the function y = f (x) is an even function on R, when x ≥ 0, f (x) = x ^ 2-2x-3, the expression of y = f (x) is written in the form of piecewise function

If f (x) is an even function on R, then f (x) = f (- x)
When x > = 0, f (x) = x ^ 2-2x-3
When x0, f (x) = x ^ 2-2x-3
When x = 0, f (x) = - 3
When x

Let f (1-x / 1 = x) = x, then the expression of F (x) is
Is f (1-x / 1 + x) = X

It's "f (1-x / 1 + x) = x," right
Let a = (1-x) / (1 + x)
Then a + 1 = (1-x) / (1 + x) + 1 = (1-x + 1 + x) / (1 + x) = 2 / (1 + x)
1+x=2/(a+1)
x=2/(a+1)-1=(2-a-1)/(a+1)=(1-a)/(a+1)
So f (a) = (1-A) / (a + 1)
So f (x) = (1-x) / (1 + x)

With the sign Max {Y1, Y2 , yn} denotes functions Y1, Y2 Then the minimum value of F (x) = max {1-x, (x / 2) - 5, (2x / 3) - 6} is

In the coordinate plane, make three straight lines: Y1 = 1-x, y2 = x / 2-5, Y3 = 2x / 3-6
Their intersection points are as follows:
Y1y2 intersects (4, - 3),
The intersection of y1y3 is (4.2, - 3.2),
Y2Y3 intersection with (6, - 2)
So f (x) can be divided into two parts
x

Find the minimum value of function f (x) = x2 ax + A / 2 (a > 0) on 0 ≤ x ≤ 1

First formula, then discussion, number shape combination
f(x)=x2-ax+a/2,
=(x²-ax+a²/4)+a/2-a²/4
=(x-a/2)²+a/2-a²/4
∵a>0
At 02:00,
When x = 1, f (x) has a minimum value of 1-A / 2

Find the minimum value of the function y = x2 + 1 / (x2-4) (x > 2), and find the value of X when the function takes the minimum value

Using the substitution method, take M = x ^ 2 - 4
Then x ^ 2 = m + 4
So y = x ^ 2 + 1 / (x ^ 2 - 4)
= m + 4 + 1/m
= (m + 1/m) + 4
>= 2 + 4 = 6
Take the equal sign if and only if M = 1 / m
In this case, m ^ 2 = 1, that is, x ^ 2 - 4 = 1
Because x > 2, x = root 5
That is, when x = root 5, y has a minimum value of 6

The minimum value of the function y = (AX ^ 2 + X + 1) \ \ (x + 1), (x ≥ 3 and a > 0) is 3Q

Let x = (x + 1) = (x + 1) and (x + 2) = (x + 1) + (f) be the extremum

If we know the first-order function and 2F (1) + 3f (2) = 3,2f (- 1) - f (0) = - 1, we can find the analytic expression of F (x)

Let the analytic expression of a function be y = KX + B
2f(1)+3f(2)=3
2(k+b)+3(2k+b)=3 ①
2f(-1)-f(0)=-1
2(-k+b)-b=-1 ②
The solution is obtained from (1) and (2)
k=4/9 b=-1/9
So the analytic expression of F (x) is f (x) = 4x / 9-1 / 9

If f (x) is a quadratic function and 3f (x + 1) - 2F (x-1) = 2x + 17

Let f (x) = ax & # 178; + BX + C 〈 3f (x + 1) - 2F (x-1) = 3A (x + 1) &# 178; + 3B (x + 1) + 3c-2a (x-1) &# 178; - 2b (x-1) - 2C = ax & # 178; + (10a + b) x + A + B + C 〈 a = 010a + B = 2A + B + C = 17 〉 a = 0b = 2C = 15 〉 f (x) = 2x + 15 if f (x) is a function of degree and 3f (x + 1) - 2F (x-1)

Given that f (x) is a quadratic function and satisfies f (x + 1) - 2F (x-1) = x ^ 2-2x + 17, find f (x)
As fast as you can, wait online!

F (x) = ax & sup2; + BX + CF (x + 1) = a (x + 1) & sup2; + B (x + 1) + C = ax & sup2; + (2a + b) x + (a + B + C) f (x-1) = a (x-1) & sup2; + B (x-1) + C = ax & sup2; + (- 2A + b) x + (a-b + C) so [ax & sup2; + (2a + b) x + (a + B + C)] - 2 [ax & sup2; + (- 2A + b) x + (a-b + C)] = x & sup2; - 2x + 17-a