The function f x ax 3 6ax 2 b, x 1,2 has a maximum value of 3 and a minimum value of 29 and finds a

f 3ax(x-4)=0,x=0 [-1,2],x=4 does not belong to [-1,2] and is therefore discarded.

1 x<0, f >0, f(x) are increments.

At 00, f(-1)>f(2), f min=f(2)=-16a+3=-29, a=2

When a<0, f(-1) so. a＝2，b＝3。

From the meaning of the title, it can be seen that a≠0 (at this time f(x)=b, f(x)min=f(x)max).

f'(x)=3ax 2-12ax=3ax(x-4) if f'(x)=0, then x=0 or x=4 when a 0, 1 x 0, then f'(x)＞0

x=0, then f'(x)=0

0 x 2, then f'(x)＜0

Therefore, f(x) is also the maximum at x=0, so f(x)max=f(0)=b=3

f(-1)=-7a+b f(2)=-16a+b at this time: f(2) f(-1).

f(x)min=f(2)=b-16a=-29 a=2

Similarly, when a 0 has -1 x 0, then f'(x)＜0

x=0, then f'(x)=0

0 x 2, then f'(x）＞0

In this case, f(x) is also the minimum at x=0, and f(x)min=f(0)=b=-29

f(-1)=-7a+b f(2)=-16a+b at this time: f(-1) f(2).

So f(x)max=f(2)=b-16a=-29 a=-2

So a=2, b=3 or a=-2, b=-29

f 3ax(x-4)=0,x=0 [-1,2],x=4 does not belong to [-1,2] and is therefore discarded.

1 x<0, f >0, f(x) are increments.

At 00, f(-1)>f(2),f

min=f(2)=-16a+3=-29,a=2.

When a<0, f(-1)min=f(-1)=-7a+3=-29, a=32 7No solution.

So. a＝2，b＝3。

f Shirt with bright or wide row vertical 3ax(x-4)=0,x=0 [-1,2],x=4 does not belong to [-1,2] so it is discarded.

1 x0, f(x) is the increment function.

0f(2),f min=f(2)=-16a+3=-29,a=2.When a

f(x) and f'The monotonicity of (x) is the same, right? (It's a long time ago, and there is a mess and some of it is forgotten--!.)

If it's the same, just look at my burning base answer, it's not the same, just ignore it.

Solution: 1) First of all, a cannot be equal to 0, otherwise f(x)=b has no maximum or minimum value in [-1,2].

2) Derivative: f'(x)=3ax^2-12ax=3a(x-2)^2-12a,f'(x) is monotonic at [-1,2] (f(x) and f are involved here'The question of whether the monotonicity of (x) is the same, I will do it in accordance with the following).

3) If a>0, then f(-1)=3, f(2)=-29 ==a=32 9, b=251 9 (-can't you make a divisible number).

If a a = -32 9, b = -485 9

f(x)=ax^3-6ax^2+b,x∈[-1,2]f'(x) = 3ax^2 - 12ax = 3a(x+2)(x-2)

Assuming a 0, the function is monotonically decreasing in the interval [-1,2], f(-1)=3, f(2)=-29

a-6a+b=3,8a-24a+b=-29, i.e.: -7a+b=3,-16a+b=-29

a=32/9，b=251/9

Assuming a 0, the function increases monotonically in the interval [-1,2], f(-1)=-29, f(2)=3

a-6a+b=-29,8a-24a+b=3, i.e.: -7a+b=-29,-16a+b=3

a=-32/9，b=-37/9

f 3ax(x-4)=0,x=0 [-1,2],x=4 does not belong to [-1,2] and is therefore discarded.

1 x<0, f >0, f(x) are increments.

At 00, f(-1)>f(2), f min=f(2)=-16a+3=-29, a=2

When a<0, f(-1) so a2, b 3.

a<>0

f'(x)=3ax 2-12ax=3ax(x-4)=0 x1=0 x2=4 (not in the interval).

f(-1)=-7a+b f(0)=b f(2)=-16a+b

1) If a<0, the maximum value = -16a + b = 3, and the minimum value ==

9-4(a-y)(b-y)≥0

Talk about reasoning in the whole thing, yes.

y²-(a+b)y+ab-

1/2≤y≤11/2

Construct inequalities.

y-1 2)(y-11 2) sells digging 0

I tidy it up and get it. y²-6y

This unequal dispersion is the same inequality as inequality (1).

a+b=6ab-9/4=11/4

Get a+b=6

ab=5a, b are the two roots of the equation m -6m + 5=0.

m-2)(m-3)=0

m = 2 or m = 3

a = 2 b = 3 or a = 3 b = 2

From the table, it can be seen that when x=0 f(x) obtains the maximum value b=3 and f (0)=-29, then f(2) f(0), which is impossible, f(2)=8a-24a+3=-16a+3=-29, a=2 if a 0, the same can be obtained a=-2, b=-29

So the answer is: a=2, b=3 or a=-2, b=-29

f(x)=ax -2ax+3-b, the axis of symmetry is x=1a>0 opening upwards.

Therefore, fmin=f(1)ruined finger=a-2a+3-b=2fmax=f(3)reputation=9a-6a+3-b=5 to get a=3 4 b=1 Qingyu talk 4

f(x)=ax²-2ax+3-b

a(x-1) 2+3-b-a monotonically increases at x 1 and f(x) has a maximum value of 5 and a minimum value of 2 at [1,3].

So f(1)=2 f(3)=5

3-b-a=2 3a+3-b=5

So a=3 4 b=1 4

f(x)=ax²-2ax+3-b

a(x^2-2x+1)+(3-b)/a -a=a(x-1)^2+(3-b)/a -a

a>0

When x=1, f(x) has a minimum value of 2

i.e. (3-b) a -a = 2

When x=3, f(x) has a maximum value of 5

i.e. 3a + (3-b) a = 5

The solution yields a=3 4 b=15 16

Because f(x)=ax -2ax+3-b=a(x-1) 2+3-a-b and because a>0, f(x) is an increasing function on [1,3], so the minimum value of f(x) on [1,3] is 3-b-a, and the maximum value is 3+3a-b

From the question, we know that 3-b-a=2 and 3+3a-b=5

The solution yields a=3 4, b=1 4