If the square 1 of the function f x 4x x is in the interval m, and 2m 1 is an increasing function, t

Solution: f(-x)=-f(x), f(x) is an odd function on r, so only the monotonicity of x 0 needs to be examined.

When x>0, f(x)=4x(x2+1)=4 (x+1x)=4 [(x-1x)2+2].

Obviously, when x>1, x>1 x, the denominator is greater than 0 and increases with the increase of x, so f(x) monotonically decreases;

When 0 is x=0, f(x)=0. Therefore, f(x) is monotonically increasing on [0,1].

Considering the symmetry of the odd function, the corresponding interval on r- is still an increasing interval in r+. Therefore, f(x) also increases monotonically on [-1,0].

So the monotonically increasing interval of the function f(x) is [-1,1].

The interval (m,2m+1) is a monotonically increasing function, so only .

1≤m≤11≤2m+1≤1

m<2m+1

The solution is -1 and I hope to be satisfied!

f(x)=4x (x2+1) is an odd function.

When x>0.

f(x)=4/(x+1/x)≤2

There is a maximum value if and only if x=1.

Therefore, the single increase interval is (-1,1).

So -1 m<2m+1 1

1≤m≤0

The axis of symmetry is x=-m

xm.

When x=-, f(-m)=1-m2 is minimal.

f(-1)=2-2m

f(2)=5+4m

5+4m=4m=4m=4 m=-1 4

When 2-2m=4, m=-1

m=-1, f(2)=9>4 does not spine stool god quarrel.

So m=-1 Sakura Blind Brigade 4

f(x)=4x (x2+1) is an odd function.

When x>0.

f(x)=4/(x+1/x)≤2

There is a maximum value if and only if x=1.

Therefore, the single increase interval is (-1,1).

So -1 m<2m+1 1

1≤m≤0

Solution 1: From the derivative, on [-1,1] is an increasing function, so (m,2m+1) is a subset of it.

So (-1,0]

Solution 2: Classification: (1) When x=0, the function is 0

2) x is not 0, the numerator and denominator are x, so y=1 (x+1 x), you write it on scratch paper, so g(x)=x+1 x, you try to draw an image, simple, and then pour it down.

The following proves that f(x) is an increasing function on (1,1).

Take -1, then f(x1)-f(x2)=4x1 (x1 2+1)-4x2 (x2 2+1)=(x2-x1)(x1x2-1) (x1 2+1)(x2 2+1)<0

The problem is proven.

f(x) is obviously an odd function.

by inequality 2|ab|<=a2+b2, and take the equal sign when a=b, or a=-b.

Result: -2=<4x (x 2+1)<=2, when x=-1, f(x) has a minimum value of -2, and when x=1, f(x) has a maximum value of 2

This increases the interval (-1,1), so we have: -1 = solution: -1

m<2m+1 gives m>-1, so if x>0, f(x)>0, let g(x)=1 f(x), when f(x) is an increasing function, g(x) is a decreasing function, g(x)=1 4(x+1 x), and its subtraction interval is (0,1], m>=0 and 2m+1<=1, and m=0

When x<0, f(x)<0, let g(x)=1 f(x), when f(x) is the increasing function, g(x) is the increasing function, g(x)=1 4(x+1 x), and its increasing interval is (-infinity, -1], so that 2m+1<=-1, which is contradictory to m>-1, so m=0

It seems that there is nothing wrong with the process, but if you use the derivation, it will be much simpler, I hope it will help you.

Since the quadratic coefficient of the function is 4 > 0, it can be concluded that the function is an open-opening upward parabola, and the axis of symmetry is x = m (2*4) =2 (there is a minimum value of the function, and the monotonic interval changes around the minimum value), and the solution gives m = 16 .

So, the original formula of the function is f(x)=4x 2 + 16x + 5

Solution: f(-x)=-f(x), f(x) is an odd function on r, so only the monotonicity of x 0 needs to be examined.

When x>0, f(x)=4x(x2+1)=4 (x+1x)=4 [(x-1x)2+2].

Obviously, when x>1, x>1 x, the denominator is greater than 0 and increases with the increase of x, so f(x) monotonically decreases;

When 0 is x=0, Yamamori slag, f(x)=0. Therefore, f(x) is monotonically increasing on [0,1].

Considering the symmetry of the odd function, the increasing interval on r+ is quiet, and the corresponding interval on r- is still an increasing interval. Therefore, f(x) also increases monotonically on [-1,0].

So the monotonically increasing interval of the function f(x) is [-1,1].

The interval (m,2m+1) is a monotonically increasing function spring balance, so only .

1≤m≤11≤2m+1≤1

m<2m+1

Solution-1

f(x)=x-2(1-m)x+2 is in the decreasing interval (1-m) on r, and to make the function a subtraction on (4), then (4) is contained in (1-m), i.e., 4 1-m, m -3

Method: Draw a picture of the quadratic function and observe the position of the axis of symmetry.

Because, the function f(x)=xsquared -2(1-m)x+2 in the interval (-infinity, 4) is a subtractive function.

Therefore, the intersection of the axis of symmetry y=1-m and the x-axis of the function is to the right of x=4, i.e., 1-m 4

So m -3

The axis of symmetry is x=-m

Decreasing at x<=m and increasing at x>m.

When x=-, f(-m)=1-m2 is minimal.

f(-1)=2-2m

f(2)=5+4m

5+4m=4m=4m=4 m=-1 4

When 2-2m=4, m=-1

When m=-1, f(2)=9>4 is incompatible.

So m=-1 4

Analysis: From the meaning of the question: the axis of symmetry x=-m, compare the axis of symmetry with the value of the midpoint of a given interval:

m<=(1+2) 2==>m>=-1 2, max=f(2)= 5+4m=4==>m=-1 4

m>1 2==>m<-1 2, max=f(-1)=2-2m=4==>m=-1

The value of the real number m is -1 or -1 4