The sufficient and necessary condition is that the maximum point of continuous function f (x) on interval [a, b] is the point where the function obtains the maximum value in this interval? In addition, if it is changed to the minimum value point and minimum value, the conclusion should be the same? When learning the binomial theorem, there is a kind of problem which is to find the term with the largest coefficient. Let PN denote the coefficient, then the standard solution is the n-value satisfying the equations PN > PN + 1, PN > PN-1. But the N solved by this formula represents the maximum point of function f (n) = PN, The maximum is not necessarily the maximum, but the standard answer here directly regards the maximum as the maximum. When are the two equal?

The sufficient and necessary condition is that the maximum point of continuous function f (x) on interval [a, b] is the point where the function obtains the maximum value in this interval? In addition, if it is changed to the minimum value point and minimum value, the conclusion should be the same? When learning the binomial theorem, there is a kind of problem which is to find the term with the largest coefficient. Let PN denote the coefficient, then the standard solution is the n-value satisfying the equations PN > PN + 1, PN > PN-1. But the N solved by this formula represents the maximum point of function f (n) = PN, The maximum is not necessarily the maximum, but the standard answer here directly regards the maximum as the maximum. When are the two equal?

There are several necessary and sufficient conditions
(1) When the continuous function f (x) has only one maximum point in the interval [a, b], if the maximum value is greater than the values of both ends f (a), f (b), then it is true
(2) When the continuous function f (x) has more than one maximum point in the interval [a, b], it must satisfy that the maximum point is the largest of all the maximum points in the interval, and is larger than the values of the endpoints f (a), f (b)
Change to minimum, minimum also hold water
In binomial theorem, binomial coefficient has only one maximum and no minimum (this conclusion can be proved, which is more complex), then the maximum is the maximum, which can be verified by drawing function image

Find the symmetry axis and vertex coordinates of the function image of y = - 2x ② + X-1, and draw the image
I want to add the value range of YX
Can you write the value coordinates on the function image?

The axis of symmetry of this parabola is x = - B / (2a) = 1 / 4,
Form vertex type:
y=-2( x@- (1/2)x)-1
=-2(x+1/4)@-7/8
Therefore, the vertex is (1 / 4, - 7 / 8). This image has a downward opening, does not intersect with the X axis, and intersects with the Y axis at the point (0, - 1) below the X axis. The image is not easy to draw. The value of X is all real numbers, and the value of Y is y less than or equal to (- 7 / 8)

Given the function y = x & # 178; - 2x - 1, the analytic expression is written in the form of piecewise function

When X & # 178; - 2x > = 0
That is, x = 2
y=x²-2x-1
When X & #178; - 2x

Let the following functions be in the form of y = a (X-H) &# 178; + K, 1, y = x & # 178; + 6x + 102, y = - 2x & # 178; - 5x + 7. The first one is in the form of y = a (X-H) &# 178; + K

1.
y=x²+6x+10
=x²+6x+9+1
=(x+3)²+1
two
y=-2x²-5x+7=-2(x²+5x/2)+7
=-2(x²+5/2+25/4)+7+25/2
=-2(x+5/4)²+39/2

By using the collocation method, the following analytic expressions are rewritten as y = a (x + m) ² + K
（1)y=x²-4x.（2）y=x²+3x+2.
（3）y=-x²+6x-1.（4）y=1-4x-2x².
（5）y=-1/3x²+2x+3.（6）y=1/2=1/3x²-2x.

（1)y=x²-4x.=x²－4x+4－4.=（x－2）²－4.（2）y=x²+3x+2.＝x²+3x+2.25＋2－2.25.＝（x－1.5）²－0.25（3）y=-x²+6x-1.＝－（x²－6x＋9）－1＋9.＝－（x－3）²＋8（4...

The function f (x) = | x-m | + 2m is known. (I) if f (x) is an even function, find the value of M; (II) if f (x) ≥ 2 is constant for all x ∈ R, try to find the value range of M

For X ∈ R, f (- x) = f (x) (2 points) | - x-m | + 2m = | - x-m | + 2m, | - M = 0 (4 points) (II) ∵ f (x) = | x − m | + 2m = x + m,, X ≥ m − x + 3mx < m The function f (x) decreases on (- ∞, m) and decreases on [M, + ∞]

If f (x + 1) = x2-3x-1, then f (x)=

f(x+1)=x^2-3x-1
=(x+1)^2-5(x+1)+3
therefore
f(x)=x^2-5x+3

If f (x + 1) = x2-3x + 2, then f (2x-1) = 4x2-14x + 12

I am a senior three graduates, just finished the college entrance examination, mathematics 130, have the opportunity to continue to exchange
The key to this problem is to find out how much f (x) = right?
If f (x + 1) = x ^ 2-3x + 2 and T = x + 1, then x = T-1, that is, f (T) = (t-1) ^ 2-3 (t-1) + 2 = T ^ 2-5t + 6
Then f (2x - 1), we can change t into 2x - 1
so
f(2x-1)=(2x-1)^2-5(2x-1)+6=4x^2-4x+1-10x+5+6=4x^2-14x+12.

Known 2x2-3x

2x2-3x

Given the function f (x + 2) = x2 + 3x, find f (x)

Let t = x + 2, then x = T-2
Substituting into the equation: F (T) = (T-2) ^ 2 + 3 (T-2)
f(t)=t^2-4t+4+3t-6=t^2-t-2
So f (x) = x ^ 2-x-2