f x 4x2 4mx m2 2m 2, m is the real number in the interval 0,2 , the minimum value on is 5, find the

Solution: f(x)=4x -4mx+m -2m+2(2x-m) -2m+2

If m 2<=0, i.e. m<=0, then f(0) is the minimum value.

f(0)=m²-2m+2=5

m²-2m-3=0

m-3)(m+1)=0

m1 = -1, m2 = 3 (does not conform to m< = 0, rounded) If m 2 > = 2, i.e. m> = 4, then f(2) is the minimum value.

f(2)=m²-10m+18=5

m²-10m+13=0

m3 = 5 + 2 root number 3, m4 = 5-2 root number 3 (does not conform to m> = 4, round) if 0f (m 2) = 2-2m = 5

2m= -3

m = -3 2, does not conform to 0 so m has two values, -1 and 5+2 3

I wish you progress in your studies and the title of the gold list.

f(x)=4x2-4mx+m2-2m+2=4(x-m/2)^2+2-2m

When x=m2.

f(x)min=2-2m

If 2-2m=5 m=-3 2

x=-3 4 is not in the interval [0,2].

Instructions 2-2m<5 m>-3 2

Since x>-3 4 f(x) is an increasing function.

So when x=0, f(x) minimum=5

That is, f(0)=m2-2m+2=5

m = 3 or -1

Analysis: If m=0, then the maximum value of the function f(x)=1 in the interval [-2,2] cannot be orange year4, so m≠0 so the symmetry axis of f(x) is x=-1, and the vertex coordinates are (-1,1-m), and it is obvious that the vertex abscissa is in the interval [-2,2] (3 points) (1) If the circle returns to m 0, then the function image opening is downward, when x=-1....

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)=x 2-2mx+m+1=(x-m) 2-m2+m+1, axis of symmetry x=m, opening up.

When m 0, [0,1] is monotonously increased on the right side of the axis of symmetry, and the most Yunzhou with side smile small value f(0)=m+1=-2,m=-3;

When 0 m 1, the symmetry trace Zen axis is in the interval [0,1], the minimum value = extreme value = -m 2 + m + 1 = -2, m = (1 root number 13) 2, which does not meet the condition of 0 m 1 and has no solution;

When m 1, [0,1] is on the left side of the axis of symmetry, monotonically decreasing, and the minimum value f(1)=1-2m+m+1=-2, m=4

In summary: m=-3, or 4

Seeking Branches: f'(x)=2mx+2m=2m(x+1)=0, the solution is x=-1 or m=0 (rounding) f(x) there is only one extreme value, and the maximum value is 4,-1 substituted into f(x) to find m=-3

When m=0, f(x)=1, (rounded).

When m>0, f(x) is a quadratic function with the opening pointing upward.

The axis of symmetry x = -1, in [-2,2], f(2)>f(-2), and [-2,-1] monotonically decreasing, (-1,2] monotonically increasing, so, f(max) = f(2) = 4m + 4m + 1 = 4

m=3 8 (satisfied).

m<0, f(x) is the function of the front cover of the two-shirt with the opening downward, the axis of symmetry x=-1, in [-2,2] or noisy, f(2)m=-3 (satisfied).

1)f'(x) = 4x 3 + 2mx because f'(2)=24 The answer line is solved with 4x 3+2mx=24 to obtain m=-2(2), and f(x)=x 4-2x 2+5 can be obtained from (1).f'(x) = 4x 3-4x let f'(x)=4x 3-4x=0 x1=0; x2=-1;x3=1。So x1; x2;x3 is the pole of the function because f(-2)=13;f(-1)=4;f(0)=5;f(1)=4;f(2)=13, so the maximum value is 13; The minimum value for clearing is 4

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

The axis of symmetry of the function is x=m

When 0 m 1 contains the axis of symmetry, the maximum value of the function is the value of the function at x=m.

m²+2m²+1-m=2

m²-m-1=0，m1=（1+√5）/2，m2=（1-√5）/2。At this time, it is not in line with the friendship 0 m 1, and it is discarded.

When m 0, the interval does not contain the axis of symmetry, and the maximum value of the function is the value of the function corresponding to the value of the x value closest to m.

x=0，1-m=2，m=-1

When m 1, the interval does not contain the axis of symmetry, and the maximum field balance of the function is the value of the function corresponding to the value of x closest to 1 m.

x=1，-1+2m+1-m=2，m=2

In summary, m=-1 or m=2

In fact, this question is still very simple:

First of all: Transform the original form into the following formula (there are two of them!). Primitive = -3x 2+1)m+4(x-1 4) 2-1 4 Because m belongs to r, then the maximum value of the previous term is equal to 0, and the last term is to obtain the minimum value when x=1 4.

And in the range from 0 to 1, the maximum value is obtained because 1 is the furthest away from 1 4.

Final Answer: When x=1, f(x)max=2