If the function y mx 2 1 m 2 x 3 is an even function and is a subtraction function on 2,3, then m

m=0, y=x-3

Not satisfied. m≠0.

f（x）=f（-x）

solution, m=+-1

m=1, y=x 2-3=(x+1) 2-4, increasing from 1 to positive infinity, does not match.

m=-1, y=-x2-3=-(x-2) 2-2, which is in line with the title, m=-1 In summary, m=-1

The title has been changed.

y=mx 2+(1-m2)x-3 is an even function.

When m=0 is y=x-3 is not an even function, it doesn't fit the topic.

When m is not equal to 0.

mx^2+(1-m^2)x-3=mx^2-(1-m^2)x-32(1-m^2)x=0

1-m^2=0

m=1 m=-1

y=x^2-3 huo y=-x^2-3

Subtraction function on the function zai [2,3].

y=x 2-3 opening up Increment on [2,3] does not match the title.

So y=--x 2-3

m=-1

Right? This can't be an even function. The axis of symmetry is x=-(1+m 2) m, if it is an even function, x=-(1+m 2) m=0, but this is not possible, unless it is an imaginary number, but it should not, so please check the question.

If it's wrong, you can correct it and I'll answer it for you, thank you.

The power function f(x)=x (m 2-2m-3)(m z) is a monotonic subtraction function over the interval (0,+ repatriation;

So m 2-2m-3 <0;

The solution is -1m z so m

m=1, m2-2m-3 =-4, so f(x)=x (1 2-2*1-3)=x(-4);

In this case, f(x) is an even function, and the world bureau decreases on (0,+ and increases on (- 0).

m, m2-2m-3=-3, so f(x)= x(-3);

At this point, f(x) is the odd spring noise function, decreasing on (0,+) and decreasing on (- 0).

(1) The primary function must be defined as y=ax+b(a≠0), so as long as m+2≠0 can make the function a primary function.

m≠22) proportional function definition: y=kx(k≠0), so as long as m+2≠0,3-3m=0 can make the function a one-time function to get m=1

(1) When m+2 is not equal to m, that is, m is not equal to 2, this function is a one-time function.

2) When 3-3m=0, that is, m=1, this function is proportional.

y=-2x 2+mx-3=-2(x-m 4) 2-3+mm 8 function is subtractive at (m 4, +00).

to make it a subtractive function on [-1, positive infinity).

m 4 < = -1, i.e. m < = -4

The axis of symmetry of the function x=-b 2a=m 4, which is a subtractive function on [-1, positive infinity), i.e., m 4<=-1, m<=-4

f(-x)=-2 3x 2-(3m-2)x-1 Spring return contains 2m-1, because f(x) is an even function, so f(x)=f(-x) is ridiculed with -2 3x 2+(3m-2)x-1 2m-1=-2 3x 2-(3m-2)x-1 2m-1

That is, Shiju (3m-2) = -3m-2) = 0

The solution is m=2 3

The parabola opens upwards.

The axis of symmetry x=-b 2a=-m

In (-2) is the subtraction function.

To the left of the axis of symmetry is the subtractive function.

It can be seen that the axis of symmetry moves with the image of the function.

Therefore, the axis of symmetry should be greater than or equal to 2

i.e. -m 2m -2

The most important thing about this kind of problem is the combination of numbers and shapes, and if you can't think of it at once, you can draw a picture to help you understand it.

f(-x)=f(x)

Then (m-1)(-x) +2m(-x)+3=(m-1)x +2mx+3(m-1)x -2mx+3=(m-1)x +2mx+3, i.e. -2mx=2mx

4mx=0 is true for any x.

So m=0 or as you say.

The even function is symmetrical with respect to the y-axis.

The y-axis is x=0

So the axis of symmetry is x=0

i.e. -2m2(m-1)=0

So m=0

The even function is symmetrical with respect to the y-axis.

That is, the axis of symmetry is x=0

So -2m2(m-1)=0

So m=0