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Answer 1 · 2014-11-08T01:07:57

#h(t)=t^{3/4}-3 t^{1/4}#

Note that the domain of #h# is #[0, infty]#.

By differentiating,

#h'(t)=3/4 t^{-1/4}-3/4t^{-3/4}#

by factoring out #3/4#,

#=3/4(1/t^{1/4}-1/t^{3/4})#

by finding the common denominator,

#=3/4{t^{1/2}-1}/{t^{3/4}}=0#

#=> t^{1/2}=1 => t=1#

Since #h'(0)# is undefined, #t=0# is also a critical number.

Hence, the critical numbers are #t=0, 1#.

I hope that this was helpful.