Ask a high school math problem about evaluation ranges.

5+4cosx=5(sinx^2+cosx^2)+4cosx=5sinx^2+5cosx^2+4cosx=cosx^2(5/tanx^2+5+4/cosx)

f(x)=tanx/√(5tanx^2+5+4/cosx)=tanx/√(5(tanx)^2+5+4√[1+(tanx)^2]

Let tanx=t

f(t)=t [5t 2+5+4 (1+t 2)] if and only if t=1 [5t 2+5+4 (1+t 2)].

The equal sign holds, and f(t) achieves the maximum.

1/t^2=5t^2+5+4√(1+t^2)f(t)<=2*1/4=1/2

Because f(t) is an odd function.

According to symmetry.

The minimum value is f(t)=-1 2

So the range is [-1 2,1 2].

Where's the root number? How to solve the problem if you don't see it clearly?

5 of these can be divided into 4+sinx and cosx, respectively, the sum of squares.

Divide sinx into situations "0 or <0".

Putting it in the denominator is only required.

where sinx > 0 is transformed into 1 root number: (2+cosx) the square of sinx + 12+cosx) the square range of the whole sinx can be solved.

[4-(x+1) 2] is originally a number greater than 0, how can the clan companion take the negative?? It's like megavolts and x are always "0".

Think about it well o( o haha

It is impossible to take a negative value for the root number

So the range after removing the root number is [0,2].

Note that the value of y is greater than or equal to zero, and the root number is added

y=sinx (2-cosx),2y-ycosx=sinx,4y-4y cosx+y cos x=1-cos x,(1+y)cos x-4y cosx+4y -1=0, so that the above equation holds =16y 4-4(1+y)(4y-1) 0,- 3 3 y 3 3,value range [- 3 3, 3 3].

y=x+ (1-x 2),1-x =y -2yx+x ,2x -2yx+y -1=0, so that the above equation holds =4y -8y +8 0,y 2, the range [- 2, 2].

(1) Inverse solution.

2) Trigonometric substitution method, x has a range limit, and the unusable method uses the boundedness of the trigonometric function to control the y range.

Haha, only for the method, not to solve the problem, this is my principle, of course it will not be adopted, hehe!

Haha, if you have a request, you just don't solve the problem!

The first question process is right?

Multiply the denominator to the left, put the two items with sinx and cosx together, extract, the sum of the squares of the coefficients, hehe, around!

y=x+ (1-x 2)·· What the hell is this, can it be regular?

The range of y is the slope of the line where the point (cosx, sinx) and (2, 2) are located.

cosx,sinx) trajectory is a unit circle, and the line in which (2,2) (which cannot be perpendicular to the x-axis) is set to kx-y-2k+2=0

The distance from the straight line to the origin is less than or equal to 1, that is, |-2k+2|Root number (k 2+1) 1, solution (4-root number 7) 3 k (4 + root number 7) 3

So the range is [(4-root7) 3,(4+root7)3].

1 Separation method.

y=(3x+1) (x-2), the domain is defined as.

y=(3x+1) (x-2)=7 (x-2)+3When x≠2, 7 (x-2)≠0, y≠3

The range of functions is;

2. Take advantage of the monotonicity of functions and combine numbers and forms.

y=5 (2x -4x+3), the domain is defined as r, y=5 (2x -4x+3)=5 [2(x-1) +1], x r,2(x-1) +1 1;

0<5/2(x-1)²+1≤5

The function value range is (0,5).

3 Discriminative method.

y=(2x -x+2) (x +x+1), the domain is defined as r, and y=(2x -x+2) (x +x+1) can be reduced to y(x +x+1) = 2x -x+2

y-2)x +(y+1)x+(y-2)=0, when x r, the above equation has a solution.

When y=2, x=0, y=2;

When y≠2, y+1) 4(y-2) 0

3y-3)(-y+5) ≥0

y-1)(y-5) 0, gives 1 y 5, and y≠2, 1 y 5 and y≠2, and the domain of the function is defined by , which is [1,5].

y=(3x+1)/(x-2)

3(x-2)+7]/(x-2)

3+7/(x-2)

y≠3y=5/(2x^2-4x+3)

5/[2(x-1)^2+1]

2(x-1)^2+1>=1

00 so define the domain as r

into a quadratic equation with respect to x.

y-2)x 2+(y+1)x+y-2=0 =(y+1) 2-4(y-2) 2>=0 to get 1<=y<=5

We have learned to copy important inequalities: (1) when a,b>0, (a+b) 2>= ab, (2) when a,b,c>0, (a+b+c) 3>= cubic root under abc above and the second problem is to use the formula (2) y=x 2(1-x), because the condition 00, 1-x >0 y=x*x*(1-x) [If you add up with the plus sign in the equation, x can't be eliminated, you can only multiply a 2 here in (1-x), and multiply it by a 1 2]=1 2*x*x*(2-2x)<=1 23rd power = 1 2 (2 3) 3rd power = 4 27 If and only if x = 2-2x, the equal sign holds, so x=2 3

Bringing in 4 and 5, the answer is the value range.

Thick, but also faint,