When x 1,2, the inequality 2x 2 mx 8 0 is constant, and the value range of m is obtained

Let f(x) = 2x +mx+8, axis of symmetry x= -m 4, and f(x) 0 when x (1,2).

It only needs to be satisfied.

f(-m 4) 0, solution m 64, m 8 or -8f(1) 0, solution m -10

f(2) 0 solution gives m -8

In summary, m -10

There are three cases: first, when -4/m is less than 1, that is, m>-4 gets 8+2m+8<0 gets m<-8

Second, when -4/m is greater than 2, that is, m<-8 gets 2+m+8<0 gets m<-10

Third, when -4/m is greater than 1 and less than 2, i.e., -8

From the known, m>8 x+2x, then it is to be constant, let f(x)=8 x+2x, then m> the maximum value of f(x), the derivative of f(x) =-8 (x 2)+2, when x (1,2) its derivative is less than 0, so it decreases in this range, so its maximum value is less than f(1)=10, so m>=10

The deformation is m<(-8-2x 2) x, only the minimum value of the polynomial on the right is required, and the derivative of y=(-8-2x 2) x is deduced and y is monotonically increased on x (1,2), so when x=1, y is min. =-10, so m<-10. Since x is an open interval, m can be given an equal sign, i.e. m is less than or equal to -10

X2+MX+4>0 Hengbi Silver was established.

That is, m is greater than the sock car - (x+4 x) constant establishment.

x+4 x is greater than or equal to 2 root number 4=4 (if and only if x=2, the equal sign holds) - (x+4 x) is less than equal to -4, i.e. m>-4

Let f(x)=x 2+mx+4

then from the known, obtained.

Only f(1) 0 and f(2) 0 are needed to draw the function image, so 1+m+4 0 is m -5

And 4+2m+4 0 is m -4

So, m -5

Slightly simplify m>(-4-x 2) x=-(x+4 x)<=2 when and modulo only when x=4 x, the equal sign Bilun into an immediate dan number of games x=2(mean inequality) so the range of m is (-5, -2).

x 2+mx+4<0 is constant at x (1,2).

m<-(x 2+4) x=-(x+4 x)-(x+4 x) in the range of x (1,2) is (-5,-4) so m can be fetched to -5, because -(x+4 x) cannot get -5, and m is always less than -(x+4 x), -5 is invariably less than -(x+4 x) (x (1,2)).

One drawing: you know that f(x)=x 2+mx+4 must pass the point (0,4) To f(x)<0 is constant, the axis of symmetry needs to be on the right side of the origin, that is.

m<0 and f(1) 0, f(2) 0, so that the two root >0 are comprehensively solved to obtain the intersection m -5

i.e. m (-5].

Because x (1,2), where x belongs to the open range, m has to go to the closed range.

The second point to mention, in the case of inaccuracy, you can substitute m=-5 into the original equation to see if it meets the meaning of the question, don't you understand?

My answer is m>1It takes advantage of the constant trump problem of a quadratic function. First look at the opening direction, because the quadratic term coefficient is greater than 0 The opening image is upward, and then consider the symmetry axis, which is determined by x=-b 2a

The axis of symmetry is m, and by the condition 0 = x = 1, m must be to the right of one, and m > 1

From the meaning of the title m (2-2x) -x -1

When x=1, m r

When 0 x 1, m(x +1) (2x-2)=1 2 [(x-1)+2 (x-1)+2] is constant on [0,1, because -1 x-1 0 has the property of a checkmark.

The function is monotonically decreasing at [0,1 so the maximum value is f(0)=-1 2, so only the maximum value of m f(x) is required, so m -1 2

In summary, the value range of m is m -1 2

From the meaning of the title, f(x)=mx +2(m+1)x+9m+4 must have the opening facing downwards and there is no zero line argument.

then m<0 and =4(m+1) -4m(9m+4)<0 then m"file training-1 2

The calculation process omits the orange of the stove...

According to the title, there are:

f(x)=x^2+mx+4

Then: f(1)<0

f(2)<0

So: 5+m<0

2m+8<0

i.e. m (-5].

There is an error in Panda Strawberry Candy.

That's all right. f(1) 0(should be changed to <=) 1+m+4 0(1+m+4<=0) m-5(m<=-5).

f(2) 0 (should be replaced by <=) 4+2m+4 0(4+2m+4<=0) m -4(m<=-4).

So, m -5

Note: Inside in parentheses are my changes.