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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 【陶哲轩实分析（上）9.6及习题-Analysis I 9.6】( a )f ( x ) = ∣ x ? 3 / 2 ∣ f(x)=|x-3/2| f(x)=∣x?3/2∣ attains minimum0 0 0 at3 / 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 ( 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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