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Suppose f(t)=(3t^3)+(5t^2)+3t-5 Find the value of t n the interval [4,8] where f(t) takes on its minimum.
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find the first derivative, set it equal to 0: f'(t) = 9t^2 + 10t + 3
9t^2 + 10t + 3 = 0 solve for t
quadratic formula?
yea
ok i get -10+-sqrt(-8) / 18
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so?
@saifoo.khan
@mariomintchev
Yes. u are correct
lol
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what else do i do to get the final answer? thats not the final answer
Now insert that value in the original equation to get the value of "y"
how would i plug in the answer i got into it? its all weird because theres no square root of 8.
?
Oh since u got imaginary roots, just check the endpoints. Plug in 4 into f(t) and 8 into f(t) whichever's smaller is ur min.
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