Question

(理)已知函数f(x)=ax³-x²+bx+c(a、b、c∈R且a≠0)在区间(-∞,0)上是增函数,在区间(0,4)上是减函数.

(1)求b的值;

(2)求a的取值范围.

(文)已知函数f(x)=ax³-x²+bx+2(a、b∈R)在区间(-∞,0)及(4,+∞)上都是增函数,在区间(0,4)上是减函数.

(1)求a、b的值;

(2)求曲线y=f(x)在x=1处的切线方程.

Answer 1

(理)解：(1)∵f′(x)=3ax²-2x+b,                                             

又f(x)在区间(-∞,0)上是增函数,在区间(0,4)上是减函数,                         

∴f′(0)=0.∴b=0.

(2)∵f(x)=ax³-x²+c,

得f′(x)=3ax²-2x.

由f′(x)=3ax²-2x=0,得

x₁=0,x₂=.                                                              

∵f(x)在区间(-∞,0)上是增函数,在区间(0,4)上是减函数,

则有a＞0且≥4.

∴0＜a≤.                                                              

(文)解：(1)∵f′(x)=3ax²-2x+b,                                                

又f(x)在区间(-∞,0)及(4,+∞)上都是增函数,在区间(0,4)上是减函数,

∴f′(0)=0,b=0.                                                            

又f′(4)=0,∴a=.                                                         

(2)∵f(x)=x³-x²+2,

得f′(x)=x²-2x.

当x=1时,f′(1)=.                                                      

此时y=f(1)=,

即切线的斜率为,切点坐标为(1,).

所求切线方程为9x+6y-16=0.