1. Determine the value of $d$ using the given y-intercept.
Given f(x)=ax3+bx2+cx+d.
Since the curve passes through (0,−1):
f(0)=a(0)3+b(0)2+c(0)+d=−1
d=−1
2. Form equations involving $a$, $b$, and $c$ using the coordinates and properties of the maximum point.
The curve passes through (−1,1):
f(−1)=1⟹a(−1)3+b(−1)2+c(−1)−1=1
−a+b−c=2— (1)
Since there is a maximum point at x=−1, the first derivative is zero (f′(−1)=0).
f′(x)=3ax2+2bx+c
f′(−1)=0⟹3a(−1)2+2b(−1)+c=0
3a−2b+c=0— (2)
3. Set up a third equation using the given area under the curve.
The area bounded by C, the x-axis, x=2 and x=3 is 431 units2. Since f(x)>0 for 2<x<3:
∫23(ax3+bx2+cx−1)dx=431
[4ax4+3bx3+2cx2−x]23=431
(481a+9b+29c−3)−(4a+38b+2c−2)=431
465a+319b+25c−1=431
465a+319b+25c=435
Multiplying by 12 to clear the denominators:
195a+76b+30c=105— (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
2a−b=2⟹b=2a−2
Substitute b into (1):
−a+(2a−2)−c=2⟹c=a−4
Substitute expressions for b and c into (3):
195a+76(2a−2)+30(a−4)=105
195a+152a−152+30a−120=105
377a−272=105
377a=377⟹a=1
Substitute a=1 back to find b and c:
b=2(1)−2=0
c=1−4=−3
(Alternatively, solving the system of linear equations directly via a Graphing Calculator yields a=1,b=0,c=−3.)
答案a=1, b=0, c=−3, d=−1