一道SAT2数学题答好了我加分 if f(x)=x³-4x²-3x+2 which of the f

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/25 17:07:24

一道SAT2数学题
答好了我加分 if f(x)=x³-4x²-3x+2 which of the following are true?
1.the function f is increasing for x≥3
2.the equation f(x)=o has two nonreal solutions
3.f(x)≥-16 for all x≥0
答案是1和3,为什么2不对?1,3又是怎么算出来的?

f(x)=x³-4x²-3x+2
f'(x)=3x²-8x-3>=0
(3x+1)(x-3)>=0
x=3时导数>=0,即为函数递增区间,即
1.the function f is increasing for x≥3,正确.
2.因为f(-1)=-1-4+3+2=0
x=-1是一实数解;
f(0)=2,f(2)=8-16-6+2=-12=0
f'(x)=3x²-8x-3>=0
(3x+1)(x-3)>=0
x在[0,3),导数0,函数递增,
所以
x=3为最小值点,此时取最小值f(3)=3³-4*3²-3*3+2=27-36-9+2=-16
从而
f(x)≥-16 for all x≥0.
正确!
答案是1和3,没错!
再问：对不起，我没学过导数，能换种说法么？SAT2是不考导数的呀。
再答： f(x)=x3-4x2-3x+2 =x3-4x2-5x+2x+2 =x(x2-4x-5)+2(x+1) =x(x+1)(x-5)+2(x+1) =(x+1)(x2-5x+2) 令 f(x)=0 有 (x+1)(x2-5x+2)=0 x=-1或x=(5+√17)/2或x=(5-√17)/2 三个实数根,所以2错! f(x)=(x+1)(x2-5x+2) x+1在x>=3 递增,x2-5x+2在x>=3也递增,所以乘积f(x)=(x+1)(x2-5x+2)在x>=3也递增! 1对! 3同上面的分析!

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