A sufficient condition for y 1 x 2 with a maximum value of 4 3 is .

This question can be used in the commutation method.

Let a=(x+2), b=(y+1), and a b=k, then the circle o can be expressed as (a-2) 2+(b-1) 2=1, and (y+1) (x+2) can be expressed as b a=k

In other words, the maximum value of (y+1) (x+2) is the maximum slope of the line b=k*a at the intersection point with the circle (a-2) 2+(b-1) 2=1 and the origin of the coordinates! (Don't look at x and y at this time, you should treat a and b as the usual x and y).

Everyone on earth knows that the slope of this straight line is the largest when it is tangent to a circle, and the formula for the distance from the point to the straight line is:

It is omitted: the center of the circle is (2,1))).

1=|1-2*k|[root number(k 2+1 2)]: k=4 3 or 0

So the maximum value of (y+1) (x+2)=b a=k is 4 3.

…Enough detail!

There are two methods for this problem, one is the slope method, the other is the parametric equation method, for convenience, I only say the parametric equation, x=2 3cosa y=2 3sina, substitute (2 3sina+1) (2 3cosa+2), let the above formula equal to k, remove the denominator, get 2 3sina+1-2 3kcosa-2k=0, use the sum angle formula is 2 (3+3k 2)sin(a-b)-2k=0, tanb=k(it doesn't matter), sin(a-b)=2k 2 (3+3k 2), sin(a-b) varies from 1 to -1, so 2 is less than or equal to 1, and k is solved.

Solution: Let x=cosa, y=sina, so (y+1) (x+2)=(sina+1) (cosa+2) is set to m, i.e., m=(sina+1) (cosa+2), so sina+1=mcosa+2m, deformed as.

sina-mcosa=2m-1, i.e. (m 2+1)sin(a+ )=2m-1, so.

2m-1|(m 2+1) 1, to solve this inequality, 0 m 4 3, so the maximum value is 4 3

Summary. It is known that y=4sin(1 2x+pie 4)y, the maximum and minimum value of y, and the range of x when y is greater than 2.

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Xie Chongzheng: (x-4) must be judged by lead 0

So when x=4, there is a maximum, then, is good.

y=-（4-4）²+5y=5

Summary. y=-x -2x-4,x [-5,10], find the maximum and minimum values.

Dear, wait a while, and sort out the answer immediately.

y=-x -2x-4,x [-5,10], find the maximum and minimum values.

Wait a minute, dear. Hurry up.

y=-x -2x-4,x [-5,10], find the maximum and minimum values.

Dear, here's the answer.

Find the increase and decrease interval to determine the maximum and minimum values.

Solution: x 2 with circle y 2 4, so let x 2cos, y 2sin , then xy 4cos 乄 sin 乄 2sin (2 乄liquid 蚂).

1 Pei buried collapse sin (2 乄) 1

The maximum value of xy 2sin (2 乄) is 2, and the minimum value is -2;

x y 2 (cos 乄 sin 乄) 2 2 sin (乄 丌 4)-1 sin (乄 丌 4) 1

The maximum value of x y 2 2 2sin (乄 丌 4) is 2 2, and the minimum value is -2 2.

This plexus is a circle with a radius of 2 and the center of the circle as the origin.

x has a maximum value of 2 and a minimum value of -2

The maximum value of y is 2 and the minimum value is -2

The maximum value of xy is 4, the minimum value is -4x+y, the maximum value is 4, and the minimum value is -4

Available Matching Methods:

y=-2x^2-4x

2(x^2+2x+1）+2

2（x+1）^2 +2.

When x 0, the maximum value is taken when x=-1, and ymax=2, and there is no minimum value.

When x [-1,3], x=-1 is just the axis of symmetry, at this time ymax=2, and the minimum value fmin=f(3)=-30

y=-2x²-4x=-2（x+1)²+2,

(1): x 0, maximumf(-1)=2, there is no minimum

-1 x 3, maximum f(-1)=2, minimum f(3)=-6

Solution: Draw the function y=a2x2a4x=a2(212 101)+2212(101) 202 When x 0 enters x=11, the maximum value is 2, and the minimum value is none. When a 1 3 is a maximum value of x = a 1 is 2, substitute x = 3 to get a minimum value of a 30.

Meet the strip fittings 0 x 1,0 y

The feasible domain is 1 triangle at the beginning of the finger domain.

The optimal solution of vertices a(0,1 2),b(0,1),c(1 2,1) is b(0,1), and the maximum value is 5

Let t = x 2+4, then t 2 = x 2+4

y=(t 2+1) t=t+1 t (t 2) first prove that y=t+1 t is in [2,+

To any 2 t1 t2

y1-y2=(t1+1/t1)-(t2+1/t2)=(t1-t2)+(t2-t1)/t1*t2=(t1-t2)[(1-1/(t1t2)]=(t1-t2)[(t1t2-1)/t1t2]

Because 2 t1 t2

So t1-t20

y1-y2