ft sqrt2t t Answers are local max at 94 74 Im not sure on

f(t) = sqrt(2-t) + t

Answers are local max at (9/4 , 7/4)

I\'m not sure on how to get the answer

to find maxima we use df/dt=0 df/dt= 1/2 * (2-t)^(-1/2) * -1 +1 (first term gets -1 from chain rule) set this equal to zero -1/2 * (2-t)^(-1/2) +1=0 multiply everything by 2*(2-t)^(1/2) -1 + 2*(2-t)^(1/2)=0 so (2-t)^(1/2)=1/2 square both sides 2-t = 1/4 t=2-1/4=7/4 the value of f at this point is 9/4

$f(t) = sqrt(2-t) + t Answers are local max at (9/4 , 7/4) I\'m not sure on how to get the answerSolution to find maxima we use df/dt=0 df/dt= 1/2 * (2-t)^(-1/2)$

ft sqrt2t t Answers are local max at 94 74 Im not sure on

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