Find the absolute maximum and absolute minimum values of f on the given interval. a) f(t)= t sqrt(36-t^2) [-1,6]

Question

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=

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10 hours 2021-10-12T13:39:36+00:00 1 Answer 0

Answers ( )

    0
    2021-10-12T13:40:57+00:00

    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 lies

    And we find this is maximum hence maximum at  (4.243,18)

    Minimum value is only when x = -1 i.e. -5.916

    b) f(t) = 2cost +sin 2t\\f'(t) = -2sint +2cos2t\\f"(t) = -2cost-4sin2t\\

    Equate I derivative to 0

    -2sint +1-2sin^2 t=0

    sint = 1/2 only satisfies I quadrant.

    So when t = pi/6 we have maximum

    Minimum is absolute mini in the interval i.e. (pi/2,0)

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