Given: `h(t)=t^(3/4)-2t^(1/4)`

Find the critical number(s) by setting the first derivative equal to zero and solving for the x value(s).

`h'(t)=(3/4)t^(-1/4)-(1/4)(2)t^(-3/4)=0`

`3/(4t^(1/4))-(1/(2t^(3/4)))=0`

`3/(4t^(1/4))=(1/(2t^(3/4)))`

`6t^(3/4)=4t^(1/4)`

`6t^(3/4)-4t^(1/4)=0`

`2t^(1/4)[3t^(1/2)-2]=0`

`t=0`

`3t^(1/2)-2=0`

`t^(1/2)=2/3`

`t=4/9`

The critical numbers are **t=0** and **t=4/9**.

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