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# Find the absolute maximum and absolute minimum values of f on the given interval.a) f(t)= t sqrt(36-t^2) [-1,6]absolute max=absolute min=b) f(t)= 2 cos t + sin 2t [0,pi/2]absolute max=absolute min=

Accepted Solution

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Answer:absolute max= (4.243,18)absolute min =(-1,-5.916)absolute max=(pi/6, 2.598)absolute min = (pi/2,0)Step-by-step explanation:a) $$f(t) = t\sqrt{36-t^2} \\$$To find max and minima in the given interval let us take log and differentiate$$log f(t) = log t + 0.5 log (36-t^2)\\Y(t) = log t + 0.5 log (36-t^2)$$It is sufficient to find max or min of Y$$y'(t) = \frac{1}{t} -\frac{t}{36-t^2} \\\\y'=0 gives\\36-t^2 -t^2 =0\\t^2 =18\\t = 4.243,-4.243$$In the given interval only 4.243 liesAnd we find this is maximum hence maximum at  (4.243,18)Minimum value is only when x = -1 i.e. -5.916b) $$f(t) = 2cost +sin 2t\\f'(t) = -2sint +2cos2t\\f"(t) = -2cost-4sin2t\\$$Equate I derivative to 0-2sint +1-2sin^2 t=0sint = 1/2 only satisfies I quadrant.So when t = pi/6 we have maximumMinimum is absolute mini in the interval i.e. (pi/2,0)