函數(shù)f(x)=x3-3x.
(1)求函數(shù)f(x)的極值;
(2)已知f(x)在[t,t+2]上是增函數(shù),求t的取值范圍;
(3)f(x)在[t,t+2]上最大值M與最小值m之差M-m為g(t),求g(t)的最小值.
分析:(1)求出f′(x)令其等于0求出駐點(diǎn),分區(qū)間討論函數(shù)的增減性求出函數(shù)極值即可;
(2)由圖象f(x)在[t,t+2]上為增函數(shù),即t+2≤-1或t≥1即可得到t的取值范圍;
(3)令f(t+2)=f(t)求出t的值,用(2)中t的取值范圍,和函數(shù)的駐點(diǎn)分6種情況求出函數(shù)的最大值與最小值相減得到g(t)的解析式,分別各段函數(shù)的最小值比較最小即可.
解答:解:(1)f'(x)=3x
2-3,令f'(x)=0,x=±1,如圖所示:
所以,f(x)
極大=f(-1)=2,f(x)
極小=f(1)=-2.
(2)f(x)在[t,t+2]上是增函數(shù),必須有t+2≤-1或t≥1,
所以t的取值范圍是(-∞,-3]∪[1,+∞).
(3)當(dāng)t≤-3時,m=f(t),M=f(t+2),g(t)=M-m=6t
2+12t+2,
令f(t+2)=f(t),6t
2+12t+2=0,
t=-1±,
當(dāng)
-3<t≤-1-時,m=f(t),M=2,g(t)=-t
3+3t+2,
當(dāng)
-1-<t≤-1時,m=f(t+2),M=2,g(t)=-t
3+6t
2-9t,
當(dāng)
-1<t≤-1+,m=-2,M=f(t),g(t)=t
3-3t+2,
當(dāng)
-1+<t≤1時,m=-2,M=f(t+2),g(t)=t
3+6t
2+9t+4,
當(dāng)t>1時,m=f(t),M=f(t+2),g(t)=6t
2+12t+2.
∴
g(t)= | 6t2+12t+2 | (t≤-3) | -t3+3t+2 | (-3<t≤-1-) | -t3-6t2-9t | (-1-<t≤-1) | t3-3t+2 | (-1<t≤-1+) | t3+6t2+9t+4 | (-1+<t≤1) | 6t2+12t+2 | (t>1) |
| |
g(t)最小值為
g(-1-)=g(-1+)=2+.
點(diǎn)評:考查學(xué)生利用導(dǎo)數(shù)求函數(shù)極值的能力,利用導(dǎo)數(shù)研究函數(shù)單調(diào)性的能力,利用導(dǎo)數(shù)求閉區(qū)間上函數(shù)最值的能力.