已知 f(x)=ax3+bx2+cx+df(x)=ax^3+bx^2+cx+d,其中 aa、bb、cc 和 dd 是常数。 方程为 y=f(x)y=f(x) 的曲线 CC 通过点 (0,−1)(0,-1) 且在 (−1,1)(-1,1) 处有一个极大值点。由 CC、x轴以及直线 x=2x=2 和 x=3x=3 所围成的面积为 314\frac{31}{4} 平方单位。已知当 2<x<32<x<3 时 f(x)>0f(x)>0,求 aa、bb、cc 和 dd 的值。

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完整解法

1

1. Determine the value of $d$ using the given y-intercept.

Given f(x)=ax3+bx2+cx+df(x) = ax^3 + bx^2 + cx + d.
Since the curve passes through (0,−1)(0, -1):
f(0)=a(0)3+b(0)2+c(0)+d=−1f(0) = a(0)^3 + b(0)^2 + c(0) + d = -1
d=−1d = -1

2. Form equations involving $a$, $b$, and $c$ using the coordinates and properties of the maximum point.

The curve passes through (−1,1)(-1, 1):

f(−1)=1  ⟹  a(−1)3+b(−1)2+c(−1)−1=1f(-1) = 1 \implies a(-1)^3 + b(-1)^2 + c(-1) - 1 = 1

−a+b−c=2— (1)-a + b - c = 2 \quad \text{--- (1)}

Since there is a maximum point at x=−1x = -1, the first derivative is zero (f′(−1)=0f'(-1) = 0).
f′(x)=3ax2+2bx+cf'(x) = 3ax^2 + 2bx + c

f′(−1)=0  ⟹  3a(−1)2+2b(−1)+c=0f'(-1) = 0 \implies 3a(-1)^2 + 2b(-1) + c = 0

3a−2b+c=0— (2)3a - 2b + c = 0 \quad \text{--- (2)}

3. Set up a third equation using the given area under the curve.

The area bounded by CC, the x-axis, x=2x=2 and x=3x=3 is 314\frac{31}{4} units2^2. Since f(x)>0f(x) > 0 for 2<x<32 < x < 3:
∫23(ax3+bx2+cx−1) dx=314\int_{2}^{3} (ax^3 + bx^2 + cx - 1) \,dx = \frac{31}{4}
[a4x4+b3x3+c2x2−x]23=314\left[ \frac{a}{4}x^4 + \frac{b}{3}x^3 + \frac{c}{2}x^2 - x \right]_{2}^{3} = \frac{31}{4}
(81a4+9b+9c2−3)−(4a+8b3+2c−2)=314\left( \frac{81a}{4} + 9b + \frac{9c}{2} - 3 \right) - \left( 4a + \frac{8b}{3} + 2c - 2 \right) = \frac{31}{4}
654a+193b+52c−1=314\frac{65}{4}a + \frac{19}{3}b + \frac{5}{2}c - 1 = \frac{31}{4}
654a+193b+52c=354\frac{65}{4}a + \frac{19}{3}b + \frac{5}{2}c = \frac{35}{4}
Multiplying by 1212 to clear the denominators:
195a+76b+30c=105— (3)195a + 76b + 30c = 105 \quad \text{--- (3)}

4. Solve the system of three linear equations simultaneously to find $a$, $b$, and $c$.

Solving equations (1), (2) and (3) simultaneously.
Adding (1) and (2):
(−a+b−c)+(3a−2b+c)=2+0(-a+b-c) + (3a-2b+c) = 2 + 0
2a−b=2  ⟹  b=2a−22a - b = 2 \implies b = 2a - 2

Substitute bb into (1):
−a+(2a−2)−c=2  ⟹  c=a−4-a + (2a - 2) - c = 2 \implies c = a - 4

Substitute expressions for bb and cc into (3):
195a+76(2a−2)+30(a−4)=105195a + 76(2a - 2) + 30(a - 4) = 105
195a+152a−152+30a−120=105195a + 152a - 152 + 30a - 120 = 105
377a−272=105377a - 272 = 105
377a=377  ⟹  a=1377a = 377 \implies a = 1

Substitute a=1a=1 back to find bb and cc:
b=2(1)−2=0b = 2(1) - 2 = 0
c=1−4=−3c = 1 - 4 = -3

(Alternatively, solving the system of linear equations directly via a Graphing Calculator yields a=1,b=0,c=−3a=1, b=0, c=-3.)

答案

a=1a = 1, b=0b = 0, c=−3c = -3, d=−1d = -1

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