Given the function f x 2x 3, if 0 2a b 1 and f 2a f b 3 , then the value range of T 3a 2 b is known

f(2a)=f(b+3)

That is, 4a-3 = 2b+3

Because 2a, 4a<2b+2

4a-3<2b +3

So there must be 4a-3 and 2b+3 as inverses to each other.

4a-3+2b+3=0

b=-2a0<2a0<2a<-2a+1

0t=3a2+b=3a2-2a=3(a-1 3) 2-1 30,1 4) in the subtraction interval of t.

So t(1 4) 5 16

Solution: f(x)=|2x-3|, f(2a) = f(b+3), that is, |4a-3|=|2b+3|．

Since 0 2a b+1, 4a 2b+2 and 4a-3 2b+3 must have 4a-3 and 2b+3 as inverses to each other

4a-3+2b+3=0, so b=-2a

From 0 2a b+1 we get 0 2a -2a+1, i.e. 0 a 1 4

t=3a2+b=3a2 -2a=3(a-1 /3 )2-1 /3 ．

This function t is a subtraction function on (0,1 4), so t(1 4) t t(0), i.e. -5 16 t 0, so the answer is (-5 16 ,0).

I've never done this before.

Because f(x+1)=2x+3, the silver is f(x+1)=2(x+1)+1, so it is known that f(x)=2x+1, and the auspicious key is f(2)=5, so choose c

f(x) = |x^2-2x-3| = |(x-1)^2 -4|Symmetrical to x = 1

The intersection of f(x) and x-axis (y = 0) is -1, 3

When x = 1, f(x) = 4

In the interval 0 < c < 4, any line of y = c intersects with f(x) at four points x = a, b, e, f, where a, b < 1,e, f > 1 f(a) = f(b) = f(e) = f(f) = c

When f(x) = c, if |(x-1)^2 -4|>0, then |(x-1)^2 -4| = (x-1)^2 -4 = c

x -1 = +/- √4+ c )

Because a < b < 1, a = 1 - 4+c).

If |(x-1)^2 -4|>0, then |(x-1)^2 -4| = 4 - x-1)^2 = c

x -1 = +/- √4-c)

Because a < b < 1, so b = 1 - 4-c).

So 2a + b = 2 - 2 (4+c) +1 - 4-c).

3 - 2√(4+c) +4-c)]

When c = 4, b = 1, 2a + b = 3 - 4 2 =

When c = 0, b = -1, 2a + b = -3

Let g(c) = 3 - 2 (4+c) +4-c)].

then g'(c) = -1/√(4+c) +1/2√(4-c)

Let g'(c) = 0, then 2 (4-c) = (4+c).

c = 12/5

g(12/5) = 3 - 10√(10)]/5 = 3 - 2√(10)

When 0 < c < 12 5, g'(c) <0, g(c) monotonically decreasing.

When 12 5 < c < 4, g'(c) >0, g(c) monotonically increasing.

So 3 - 2 (10) <= 2A+B < 3 - 4 2

f(x)=|x2-2x-3|，|x-3)(x+1)|

When x = 1, f(x) = 4

Let f(x)=4 x=1 or.

x=1- 2 or x=1+ 2

b has a range of -1a and a has a range of 1- 2, so 2a + b has a range of .

1-2√2<2a+b<-1

Look at the coordinates yourself, using the extreme value method.

1) From f(2 3) + ab = 20 3 can be obtained: a + b + ab = 82 root number (ab) < = a + b Therefore, 2 root number (ab) + ab< = 8 ab + 2 root number (ab) - 8 "positive start block = 0 (root number (ab) + 4) (root number (ab) - 2) < = 0

and ab>0, so 0 "root number (ab) < = 2 03 area.

z=(b+1) (a+1)+1 b+1=(z-1)(a+1) is a straight line with a passing point (-1, -1) and a slope of z-1.

To make a straight line pass through the above area, then 2 5

f(x) function image with an axis of symmetry is a straight line x= i.e. satisfies: f(3-x)=f(x).

The image is drawn by drawing 2x-3 and then flipping the x-axis down to top of it.

The meaning of the identity above is (as can also be seen in the graph): for any two variables, if the sum of the two variables is 3 (equivalent to the midpoint of the two variables, then the function values of the two are equal).

Since b+3>2a, in order to make the values of the two functions equal, we can first prove that when b+3 and 2a are both greater than or both are less than, the equality of the values of the two functions must lead to the contradiction of b+3=2a. So we can get b+3<, 3a> holds, so according to the symmetry there is:

b+3+3a=3 (i.e., the two variables b+3 and 3a are about left-right symmetry).

So t=3a 2-3a (3a> is a quadratic function, and a parabola can be drawn.

Solution: Because f(2a) = f(b+3).

So |4a-3|=|2b+3|

So (4a-3) 2=(2b+3) 2

So 16a 2-24a+9=4b 2+12b+9 so 16a 2-24a=4b 2+12b so 4a 2-6a = b 2+3b...... 2a-3/2)^2=(b+3/2)^2

2a-b-3)(2a+b)=0

So 2a=b+3 or 2a=-b

Discuss: When 2a=b+3.

Because t=3a2+b

3a^2+2a-3

3(a^2+2a/3)-3

3[(a+1/3)^2-1/9]-3

3[(a+1/3)^2]-10/3

So t>=-10 3

When 2a=-b.

Because t=3a2+b

3a^2-2a

3（a^2-2a/3)

3[(a-1/3)^2-1/9]

3(a-1/3)^2-1/3

So t>=-1 3

From the symmetry of the function image, we can see that f(3-x) = f(x).

Or from f(x)=|2x-3|=|3-2x|=|2(3-x)-3|=f(3-x), so that there is 2a+b+3=3

So b=-2a is substituted for t=3a2+b, and t=3a, 2-2a=3(a-1, 3), 2-1, 3, -1, 3, 3, -1, 3

There are two cases for the symmetry of the function.

1) When 2a<3 2 and b+3>3 2 have.

f(2a)=3-4a

f(b+3)=2b+3

At this time, there is b=-2a, so there is t=3a 2-2a(a<3 4) and there is a parabola to know t>=-1 3

2) When 2a>=3 2 and b+3<=3 2, b=-2a, t=3a 2-2a(a>=3 4) are introduced

t is an increasing function of a, so t>=t(3 4)=3 16