If the minimum value of three functions y = 4x + 1, y = - 2x + 4, y = x + 2 is denoted as y = f (x), the maximum value of y = f (x) is obtained It's a process

If the minimum value of three functions y = 4x + 1, y = - 2x + 4, y = x + 2 is denoted as y = f (x), the maximum value of y = f (x) is obtained It's a process

Three equations are established successively
4x+1=-2x+4,4x+1=x+2,-2x+4=x+2
x=1/2,x=1/3,x=2/3
Draw an image in a coordinate system
When x

It is known that f (x) is a quadratic function and satisfies f (2 + x) = f (2-x). If the function has two zeros x1, X2, then the value of X1 + X2 is zero

F (2 + x) = f (2-x) is equivalent to the axis of symmetry x = 2
So X1 + X1 = 2 * 2 = 4

Given that the zeros of quadratic function y = f (x) X1 = 0, X2 = 4, and the maximum value is 4, find the analytic expression of the function

According to the known conditions, let the quadratic function be y = (x-0) (x-4) + k = x ^ 2-4x + k = (X-2) ^ 2 + K-4 [K is any constant]
When x = 2, y is the maximum, y = K-4 = 4, k = 8
Then the analytic expression of quadratic function is y = (X-2) ^ 2 + 4

The monotone interval of function f (x) = x ^ 2-lnx ^ 2 is discussed, and the extremum is obtained

The domain is x ≠ 0
F (- x) = f (x), so f (x) is even
When x > 0, f (x) = x ^ 2-2lnx, f '(x) = 2x-2 / x = 2 (x ^ 2-1) / x, the minimum point x = 1, f (1) = 1
When x > 1, the function increases monotonically

(2-a)x-2/a,x

We can see that f (x) is a piecewise function. When x > = 1, f (x) = x ^ 2, monotonically increasing. If f (x) is monotone on R, then if x0 and (2-A) * 1-2 / a at the breakpoint

If the image of quadratic function y = ax ^ 2 (a ≠ 0) is shown in the figure, then the solution set of inequality ax + a > 0 is
The graph is the one with the opening up, and the vertex is at the origin.

The picture is the one with the opening up
Then a > 0
ax+a>0
ax＞-a
The solution is x ＞ - 1
The solution set of inequality ax + a > 0 is x ＞ - 1

The function y = - 3 (x + 2) & sup2; + 5's image opening direction?, symmetry axis?, vertex coordinates?

a=-3

The image of the function y = ax ^ 2 + C (a ≠ 0) is? The axis of symmetry is? The vertex is?

The graph of function y = ax & sup2; + C (a ≠ 0) is a parabola
Y = a (x-0) & sup2; + C is obtained by changing y = ax & sup2; + C into vertex form
The symmetry axis of the function y = a (x-0) & sup2; + C is x = 0, which is the Y axis; the vertex coordinates are (0, c)

The function y = (1 / 3) ax ^ 3 - (1 / 2) ax ^ Z (a is not equal to 0) is an increasing function in the interval (0,1), and the value of real number a
What is the range of a?

Y '= ax ^ 2-ax = ax (x-1), get: extreme point x = 0,1
On (0,1), y '> 0, because this interval has x (x-1)