Question

設(shè)函數(shù)f(x)=x³-kx²+x(k∈R).

(1)當(dāng)k=1時(shí),求函數(shù)f(x)的單調(diào)區(qū)間.

(2)當(dāng)k<0時(shí),求函數(shù)f(x)在[k,-k]上的最小值m和最大值M.

Answer 1

對函數(shù)f(x)=x³-kx²+x求導(dǎo)得f′(x)=3x²-2kx+1.

(1)當(dāng)k=1時(shí)f′(x)=3x²-2x+1,由Δ=4-12=-8<0可知f′(x)>0,f(x)在R上單調(diào)遞增.

(2)方法一:當(dāng)k<0時(shí),f′(x)=3x²-2kx+1,其圖象開口向上,對稱軸x=,且過點(diǎn)(0,1)

(ⅰ)當(dāng)Δ=4k²-12=4(k+)(k-)≤0,即-≤k<0時(shí),f′(x)≥0,f(x)在

 [k,-k]上單調(diào)遞增,從而當(dāng)x=k時(shí),f(x)取得最小值m=f(k)=k,當(dāng)x=-k時(shí),f(x)取得最大值M=f(-k)=-k³-k³-k=-2k³-k.

(ⅱ)當(dāng)Δ=4k²-12=4(k+)(k-)>0,即k<-時(shí),令f′(x)=3x²-2kx+1=0,解得x₁=,x₂=,注意到k<x₂<x₁<0,所以m=min, M=max.

因?yàn)閒(x₁)-f(k)=-k+x₁-k=(x₁-k)(+1)>0,所以f(x)的最小值m=f(k)=k;

因?yàn)閒(x₂)-f(-k)=-k+x₂-(-k³-k·k²-k)=(x₂+k)[(x₂-k)²+k²+1]<0,所以f(x)的最大值M=f(-k)=-2k³-k;

綜上所述,當(dāng)k<0時(shí),f(x)的最小值m=f(k)=k,最大值M=f(-k)=-2k³-k.

方法二:當(dāng)k<0時(shí),對∀x∈[k,-k],都有f(x)-f(k)=x³-kx²+x-k³+k³-k=(x²+1)(x-k)≥0,故f(x)≥f(k);

f(x)-f(-k)=x³-kx²+x+k³+k³+k=(x+k)(x²-2kx+2k²+1)=(x+k)[(x-k)²+k²+1]≤0,故f(x)≤f(-k).

又f(k)=k<0,f(-k)=-2k³-k>0,所以f(x)_max=f(-k)=-2k³-k,f(x)_min=f(k)=k.