f(0)=2,f(-2)=0,f(x)在x=-1，x=5处有极值，f(x)的导数是x的二次函数.

Question

求fx

Answer 1

f(x)=ax^3+bx^2+cx+d
f(0)=2,
d=2
f(x)=ax^3+bx^2+cx+2
f(-2)=0
0=-8a+4b-2c+2
f(x)的导数是x的二次函数.
f'(x)=3ax^2+2bx+c=0
在x=-1，x=5处有极值,代入得
3a-2b+c=0
75a+10b+c=0
解得
a=1,b=-6,c=-15
f(x)=x^3-6x^2-15x+2

Answer 2

设f'(x)=ax^2+bx+c
则
f(x)=∫f'(x)dx=∫(ax^2+bx+c)dx=ax^3/3+bx^2/2+cx+d
f(0)=d=2
因为f(x)在x=-1，x=5处有极值
则f'(-1)=a-b+c=0 1
f'(5)=25a+5b+c=0 2
2式-1式得
24a+6b=0
b=-4a 代入1式得
a-(-4a)+c=0
c=-5a
所以
f(x)=ax^3/3-4ax^2/2-5ax+2
=ax^3/3-2ax^2-5ax+2
因为
f(-2)=a*(-2)^3/3-2a(-2)^2-5a*(-2)+2
=-8a/3-8a+10a+2
=-8a/3+2a+2
=-2a/3+2=0
a=3
所以
f(x)=3x^3/3-2*3x^2-5*3x+2
=x^3-6x^2-15x+2

Answer 3

设f(x)=ax^3+bx^2+cx+d (a≠0)
x=0 f(x)=2代入，得d=2
x=-2 f(x)=0 d=2代入，得
-8a+4b-2c+2=0 (1)
f'(x)=3ax^2 +2bx+c
由函数在x=-1，x=5处有极值，得
3a(x+1)(x-5)=0
整理，得
3ax^2-12ax-15a=0
与f'(x)表达式对比，得
2b=-12a b=-6a c=-15a，代入(1)
-8a-24a+30a+2=0
2a=2
a=1
b=-6a=-6 c=-15a=-15
函数解析式为f(x)=x^3-6x^2-15x+2

Answer 4

由题意可知
f'(-1)=f'(5)=0
f'(x)为二次函数 则可设
f'(x)=a(x+1)(x-5)
=a(x²-4x-5)
f(x)=∫ a(x²-4x-5) dx
=a(x³/3-2x²-5x)+C
f(0)=C=2
f(-2)=a(-8/3-8+10)+C=0
解得
a=3 C=2
f(x)=3*(x³/3-2x²-5x)+2
=x³-6x²-15x+2

Answer 5

f(x)在x=-1，x=5处有极值
可以设f'(x)=a(x+1)(x-5)=ax^2-4ax-5a
所以f(x)=ax^3/3-2ax^2-5ax+c
f(0)=2,f(-2)=0，带入，得：a=3, c=2
f(x)=x^3-6x^2-15x+2

Answer 6

不妨设f'(x)=ax^2+bx+c,则f(x)=a/3x^3+b/2x^2+cx+d
f(x)在x=-1，x=5处有极值，所以f'(-1)=f'(5)=0又有f(0)=2,f(-2)=0
带入联立方程组可解得
a= 3 b= -12 c=-15 d=2
f(x)=x^3-6x^2-15x+2