Xét hàm số g(x) = -x2 + x + 6 với x ∈ [0, 1)Ta có:\(\eqalign{
& g'(x) = - 2x + 1 \cr
& g'(x) = 0 \Leftrightarrow x = {1 \over 2} \cr} \)\(\eqalign{
& g(0) = 6;\,\,\,g({1 \over 2}) = {{25} \over 4};\,\,\,g(1) = 6 \cr
& \mathop {\min }\limits_{x \in {\rm{[}}0,1{\rm{]}}} (x) = 6;\,\,\,\mathop {\max }\limits_{x \in {\rm{[}}0,1{\rm{]}}} (x) = {{25} \over 4} \cr} \)\(\eqalign{
& \Rightarrow 6 \le g(x) \le {{25} \over 4}\,\,\,(\forall x \in {\rm{[}}0,1{\rm{]}}) \cr
& \Rightarrow {2 \over 5} \le f(x) = {1 \over {\sqrt {g(x)} }} \le {{\sqrt 6 } \over 6} \cr} \)Vậy \(\mathop {\max}\limits_{x \in [0,1{\rm{]}}} (fx) = {{\sqrt 6 } \over 6};\,\,\,\mathop {\min }\limits_{x \in [0,1{\rm{]}}} (fx) = {2 \over 5}\)
