试题与答案

讨论y=1-x2在[-1,1]上的单调性.

题型:解答题

题目:

讨论y=
1-x2
在[-1,1]上的单调性.

答案:

此函数可以看成是由函数y=f(t)=

t
和t=1-x2 复合而成,对于f(t)在t≥0始终单调递增,

对于t=1-x2,在x∈(-∞,-0)上单调递增;在x∈[0,+∞)上单调递减,

有复合函数单调性的“同增异减”法则,可以知道:

-1≤x≤1
x∈{x|x<0}
⇒-1≤x<0,即当x∈[-1,0)时.函数y=
1-x2
是单调递增函数;

-1≤x≤1
x∈[0,1]
⇒0≤x≤1,即当x∈[0,1]时,函数y=
1-x2
是单调递减函数.

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