find all critical numbers of the function t square root 4-t, t is less than 3
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f(t)=t√(4-t), f'(t)=-t/(2√(4-t))+√(4-t). When f'=0 we solve for the critical value of t<3:

(-t+2(4-t))/(2√(4-t))=0, -t+8-2t=0, 3t=8, t=8/3 and f(8/3)=(8/3)√(4/3)=(8/3)*2/√3=16/(3√3)=16√3/9.
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