g(t) = 3√t^2 (2t-1)

g(t) = 3√t^2 (2t-1)

t= ? and t=1/ ?

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1 Answer

g(t)=∛t²(2t-1)=t^⅔(2t-1),

g'(t)=2t^⅔+(⅔t^(-⅓))(2t-1)=0 at extrema.

So 2t+⅔(2t-1)=0 (t≠0), 6t+4t-2=0, 10t=2, t=0.2.

Intercepts:

g(0)=0; g(½)=0.

g(0.2)=-0.2052 approx.

These include t=0 and 1/t=1/0.2=5.

by Top Rated User (1.1m points)

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