Help me see what s wrong with solving my freshman math problem like this?

When t=1 2, 1 t=2 is not equal to 1 3, that is, the left endpoint in the formula cannot be taken at the same time, and the right endpoint cannot be taken at the same time.

Correct solution: The minimum value is based on the basic inequality mentioned above, and the maximum value can be found as 2, and the maximum value should be calculated using the derivative method (the minimum value can also be found together, that is, the basic inequality should not be used at all) so that g(t)=t+1 t

g'(t)=1-1 t =(t+1)(t-1) t 1 2,1] up-subtracting function [1,3] up-increasing function.

The minimum value is g(1)=2

The maximum value is the greater of g(1, 2) and g(3).

Since g(1 2) = 5 2 g(3) = 10 3, the maximum value is 10 3

Therefore, the value range is [2,10 3].

This should be done with a basic inequality.

f(x)=f(x)+1/f(x)≥2

An equal sign is taken if and only if f(x)=1.

The Nike function has a maximum value when f(x) is taken as 3.

The f(x) range is [2,10 3].

The method you use to take the minimum and maximum values are different and cannot be added).

As mentioned above, although the addition of inequalities is true, the error will increase with the number of additions.

Besides, your calculation process is based on the result of taking different values for x.

1 2 t 3 is correct, but it does not mean 1 3 1 t 2, in this question you directly let f(x)=t, then f(x)=t+1 t, when t=1 2, substitute; f(x)=5 2, when t=3 enters: f(x)=10 3 so choose c

t+1 t Typical function ah minimum value 2 Ah · t=1.

The problem can be simplified to y=x+1 x, where the range of x is [1 2,3]; Find the range of y.

First of all, we should consider the monotonicity of y, which is not monotonically increasing or decreasing in the range [1 2,3], because it cannot be directly added or subtracted from a simple inequality. It can be analyzed with the function image of y=x+1 x, or it can be directly derived from y, and the minimum value of y=x+1 x(x>0) can be found [let y'=0, the value of x is 1]; This brings x=1 into y=x+1 x, resulting in a minimum value of y of 2; At this time, the answer has been locked in option b, if you want to refine it further, you can substitute x=1 2 and x=3 into y=x+1 x respectively, compare the size of the obtained y value [5 2 and 10 3], and take the maximum value as the upper limit of y; The range of y is: [2, 10 3].

That is, the range of f(x) is: [2,10 3]. Option b is correct.

1. y=x+1/x

Range (- 2]u [2, +

Monotonic interval (-1]u[1,+ monotonically increasing; [-1,0)u(0,1] monotonically decreasing; Drawing is easiest in one or three quadrants.

You can't add two variables together.

With basic inequalities.

1/t+t>=2

Valid when and precisely t=+1.

Obviously, t minimum = 2 to choose b

This function is not a composite function.

y=e^xx=log1/2(a)

is a composite function.

The function on the problem is not that the independent variable is another function, but that the constant is another function, so the law of same increase and difference subtraction of the composite function cannot be applied.

If the two functions that make up the composite function are both increasing or decreasing, the composite function is increasing, and if one increases and one decreases, the composite function decreases, which is "the same increase and different subtraction".

The mistake is that this is not a composite function. Therefore, there is no such thing as an increase or a difference.

The idea is wrong, you are not a composite function. When looking at the base of an exponential function alone, it is an increasing function.

In fact, during the college entrance examination, this kind of question is the fastest to take a random number. For example, 1 4 satisfies the question, and the other options do not contain 1 4, so choose a

Causal error, you should start from 2 and push forward, it is the condition, not from the front

The value range of the inner function of the composite function is contained in the definition domain of the outer function, the x power of t, t is the inner function, find the increasing interval of the x power of t, t 1, that is, log(, x is greater than 0, so it is a.

It is an exponential function with log1 2 a as the base (generally only the case where the base is greater than 0 and not equal to 1, this problem can get 0.)

Solution: Let the radius of the sector be rcm, and the central angle of the circle is 乄, then the circumference of the sector is 2r 乄 20, so 乄 (20 2r) state before r, then the area of the sector is clear s 乄 r 2 2

r^2+10r

r－5）^2＋25

When r 5, the sector area s maximum is 25 square centimeters, at which point 2 (rad).

Here's how I solved the problem:

1.Suppose the radius is a and the central angle of the circle is b, then.

2a+ba=20 a＞0，b＞0

2.The area of the sector is (ba2) 2

3.List the relationships between A and B.

b=(20-2a)/a=20/a-2

and because of b 0, Shi Zheng a 10

4.The sector area is.

20/a-2）a^2）/2=（20a-2a^2）/2=10a-a^2=-（a-5）^2+25

0＜a＜10

Therefore, when a = 5, the sector area is the largest, and the maximum area is 25, and the sector angle is b=2

As for what you said about your teacher's -2a b and other solutions, I don't know what it means. This is a typical minimum value finding problem.

Solution: 2r l 20

l＝20－2r

s＝lr÷2＝(20－2r)×r÷2＝－r²＋10r。

According to the quadratic function, the opening of the parabola is downward, and S obtains a maximum value at the vertex, and when r 2a b 5, the maximum value of s has a wheel field of 25 cm.

At this point, the central angle l r (20 2r) r 2

The perimeter of the Fan Town's Limb Destroyer consists of two radii, and a circular arc.

a b=, then: a 3-2*a 2-a+7=5(1)b must have an element equal to 2 in the set, and you can try one by one.

a+3=2 can get a=-1, then the elements in the set b are, the elements in the set a are, and the elements in the set a are not eligible.

a 2-2*a+2=2 can be obtained: a=0 or a=2 is brought into (1) to obtain a=2;Then the elements of the b set are { 4,5,2,6,25}a b meets the requirements.

a 3-2*a+2=2 can be obtained: a=0, a 2=2; None of the requirements are met.

A 3 + a 2 + 3 * a + 7 = 2 and (1) are combined to obtain: 3 * a 2 + 4 * a + 3 = 0 without solution.

So we get a=2

b = (3k 1) 3 6k 3, considering only the molecule, that is, the number of the form 3k 1 and 6k;

a=k 3, again, only the molecule is considered, that is, the number of the shape k, of course, a has a large range, so b is contained in a.

Answer: A contains B (B contains A).

You're right, exactly.

First of all, I think you should be abusing this inequality, the equality sign of inequality is conditional, for example, in your solution, a+b=2ab, a+b>=2*root sign (ab), ab>=1, here, only when a=1=b=1, take the equal sign;

Go down, 2a+b>=2 root number (2ab), here 2a=b is only taken when the equal sign, then you say, the above are a=b=1, how can 2a be equal to b, so here is not feasible at all, you can only know 2a+b>2 root number (2ab), remember, greater than is not the minimum value, the minimum value is greater than or equal to ah, right;

Therefore, the solution to this problem is to use only 1 inequality, remember to use 1 time;

Let y=2a+b

a+b=2ab, so 1 (2a)+1 (2b)=1, so y=(2a+b)(1 2a+1 2b), then, you can use your mean inequality here, because you will find that you can take the equal sign if certain conditions are met here;

This problem does have a certain degree of difficulty haha, you need to turn a little bit, the routine of this kind of problem is, most of them need to be converted into a bunch of equations xyzbluablua = 1, and then 1 * followed by a string of equations, and then use inequality.