陶哲轩实分析(上)9.6及习题-Analysis I 9.6

这一节是最值定理,即闭区间上连续函数一定可取到最大值和最小值。

Exercise 9.6.1

( a ) f ( x ) = ∣ x − 3 / 2 ∣ f(x)=|x-3/2| f(x)=x3/2 attains minimum 0 0 0 at 3 / 2 3/2 3/2, but can’t attain its maximum.
( b ) f ( x ) = a x , 0 < a < 1 f(x)=a^x,0f(x)=ax,0<a<1 attains it’s maximum 1 1 1 at 0 0 0, but can’t attain its minimum.
( c ) f ( x ) = { 0 , x = − 1 x , x ∈ ( − 1 , 1 ) 0 , x = 1 f(x)=\begin{cases}0, & x=-1\\x, & x∈(-1,1)\\0, & x=1\end{cases} f(x)=0,x,0,x=1x(1,1)x=1
( d ) f ( x ) = { 0 , x = 0 1 / x , x ∈ [ − 1 , 0 ) ∪ ( 0 , 1 ] f(x)=\begin{cases}0, & x=0\\1/x, & x∈[-1,0)∪(0,1]\end{cases} f(x)={0,1/x,x=0x[1,0)(0,1]

For the examples in ( a ) and ( b ), the intervals are not closed.
For the examples in ( c ) and ( d ), the functions are not continuous.

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