第1个回答 2011-03-05
1)
P(1,f(1))处的切线方程为y=3x+1
=>
切点:(1,3*1+1)即(1,4)
=>
f(1)=4
=>
a+b+c+1=4
f(x)=x^3+ax^2+bx+c
=>
f'(x)=3x^2+2ax+b
y=3x+1
=>
k=3
=>
f'(1)=3
=>
3+2a+b=3
-2极值
=>
f'(-2)=0
=>
3*4-4a+b=0
连列方程:
a+b+c+1=4
3+2a+b=3
3*4-4a+b=0
=>
a=-1,b=2,c=2
2)去掉1)中第三个方程
=>
a+b+c+1=4
3+2a+b=3
=>
2a+b=0
-2,1上单调递增
=>
-2,1上f'(x)>0
=>
-2,1上3x^2+2ax+b>0
=>
-2,1上3x^2-bx+b>0
以下略
应用题:
单价p
=>
p^2与x成反比
=>
p^2=k/x(k待定)
=>
p=50时,x=100
=>
50^2=k/100
=>
k=250000
=>
x=250000/p^2
利润y=单价*件数-成本
=p*x-C(x)
=(k/x)^(1/2)*x-1200-(2/75)*x^3
=500*x^0.5-(2/75)*x^3-1200
=>
y'=500*0.5*x^(-0.5)-(2/75)*3x^2
令y'=0
=>
x=5
且x>5时,y'<0;x<5时,y‘>0
故x=5时有最大值本回答被提问者采纳