The defined domain of f x mx 2 4x m 2 1 2 x 2 m x 1 0 is known

Note that the five major aspects of the definition domain are that the denominator of 1 is not 0, the power of 0 in the root number of 2 is greater than 0, the power of 0 in 3 0 is meaningless, the true number in the logarithm of 4 is greater than 0, and the base of the 5 logarithmic exponent is greater than 0 and ≠ 1

f(x)=(mx (2)+4x+m+2) (1 2)+(x(2)-mx+1) (0) is defined in the domain r

mx (2)+4x+m+2 0 is constant and x (2)-mx+1 0 m 0

mx^(2)+4x+m+2＞0

b (2)-4ac = 16-4m(m+4) 0, and m<-(5)-1 or m>(5)-1 is obtained

x^(2)-mx+1＞0

b^（2）-4ac=m^(2)-4<0

The solution is -2 and m ((5)-1,2).

It is solved in two steps, and finally the intersection is comprehensively solved.

First) = (mx (2)+4x+m+2) (1 2), to make it meaningful, it should meet mx (2)+4x+m+2>0, and treat m as a constant to solve a one-dimensional 2-order inequality (it should be solved, the computer is inconvenient to write the process, sorry): in the first case m=0, then the original inequality is 4x+2>0The solution is x>-1 2, because the definition domain of x is r, so m=0, which does not conform to the meaning of the topic, and is discarded; In the second case, m>0, in order to make x r, the discriminant formula must be less than 0, so there is 4*4-4*m*(m+2)<0, and m>[-1+17 (1 2)] 2; In the third case, m<0, the parabola opening will be downward, and it will be less than 0, so it does not meet the requirements, and m<0 is rounded.

Finally, (x (2)-mx+1) (0), which should be 0 to the power of 0 is meaningless, so (x (2)-mx+1) is not equal to 0, and m is not equal to plus or minus 2, and the sum of the above can be obtained that the range of m that satisfies the condition for defining the domain r is: [-1+17 (1 2)] 22 (can also be written as m> [-1+17 (1 2)] 2 and m is not equal to 2).

The definition field is r, which means that the definition field of both parts of f(x) is r, that is, the following two conditions must be met:

For x belongs to r, (mx (2)+4x+m+2) (1 2) is constant, i.e., mx (2)+4x+m+2>0 is constant.

For x belongs to r, (x (2)-mx+1) (0) is constant, i.e.

I don't know if there's something wrong with the theme of the building (x (2)-mx+1) (0), it doesn't seem to be to the power of 0, check it, my ability is limited.

But the basic idea is that since the defined domain is r, it means that the defined domain of each part of the function is r (because the definition domain is the intersection of small defined domains). In fact, the final problem is transformed into a regular problem, the problem of the establishment of one element and two constants, if you do not know how to do it in this step, it is recommended that you choose a few questions of constant establishment to practice writing.

If you want to ask for this kind of question, you will definitely take the college entrance examination, and you can also do some college entrance examination questions.

Go upstairs to do the second, 0 power there is a problem. The basic idea is the same, and it needs to be discussed by classifying m.

Unbridging roll: Let t=(mx2+8x+n) (x2+1) then 1==0, i.e.: t2-(m+n)t+mn-16<=0 (2).

It is equivalent to the inequality t-1)(t-9)<=0 (3) from (2)(3) to obtain :m+n=10 ,mn=16 and thus solve: m=n=5

Define the domain as r, then mx 2+4x+m+2 must be equal to 0x 2-mx+1 and cannot be equal to 0

That is. For MX 2+4X+M+2 opening upwards, the judgment formula is less than or equal to 0, that is, the high bucket.

m>0 megabytes.

16-4m(m+2)≤0 ②

For x 2-mx+1, <0, ie.

m²-4<0 ③

By the way.

m is the range.

Root number 5-1 m<2

3 (m+2)=18, m+2=log3 18=2+log3 2, m=log3 2 (the following is the true number).

g(x)=λ3^mx-4^x=λ[3^log3 2]^x-4^x=λ2^x-4^x=-(2^x)²+2^x

Let t=2 x,t be an increasing function, and see t as an independent variable, then the parabola g(t)=-t + t, the axis of symmetry t= 2, and the origin is a subtraction function on [0,1], then the axis of symmetry can only be on the right side of the y-axis, and the axis of symmetry is on the right side of [0,1].

So 2>=1, 2

The function is meaningful and needs to be satisfied: mx 2+4x+m+2>0, x 2-mx+1 0 (the number of non-negative congs under the root number is not 0).

From the parabolic image, when x r, the inequality must be m>0, that is, the opening is upward.

It also needs to be satisfied: 1=4 2-4m(m+2)<0, 2=m 2-4<0 to solve the first inequality, and m> 5-1 is obtained

To solve the second equation of the sock sedan, 0 takes its intersection to get 5-1

The function f(x)=(mx 2+4x+m+2)-3 4+(x 2-mx+1) 0 is defined in the domain r

x 2-mx+1≠0 is constant for any x r.

-m）^2-4*1*1＜0

The value range of 2 m 2 is (-2,2).

f(x)=-1/2x2+x=-1/2(x-1)^2+1/2<=1/2.

The opening is downward, and the axis of symmetry is x=1

The range is [2m,2n], so there is 2n<=1 2,n<=1 4 So, the interval [m,n] is on the left side of the axis of symmetry, monotonically increasing the function.

So there is: f(m)=-m2 2+m=2m

f(n)=-n^2/2+n=2n

The solution gives m=0, or m=-2, n=0, or n=-2 due to mm+n=-2

Discussion on a case-by-case basis:

1. m and n are both on the left side of the axis of symmetry, that is, m m=0 or m=-2, n=0 or n=-2, so m=-2, n=0.

m+n=-2

2. m<=12-1, simplified to 1 2(n-m)(m+n-6)=0, n-m>0, m+n+6=0, m+n=6

In summary, m+n=6 or -2

From f(m)=(1 2)m +m=2m, m(m+2)=0, m1=0, m2=-2

For the same reason, f(n)=(1 2)n +n=2n, n1=0, n2=-2

Due to the conditions, mf(0)=0=3*0Therefore, when x [-2,0], f(x) [4,0] satisfies the requirement.

f(x)=x²-2x+3

x-1)^2+2

Axis of symmetry x = 1

If m+1 1 then.

The maximum value is f(m) and the minimum value is f(m+1).

If m<12 then.

The maximum value is f(m+1) and the minimum value is 2

If m 1 then.

The maximum value is f(m+1) and the minimum value is f(m).

Find the value of x corresponding to the minimum value of f(x), and discuss the situation in which m+ is less than the value of x, m+ is greater than the value of x, and m+ is equal to the value of x.

It depends on the range of m values.