Find the minimum of g(t)=t^3-25t+6 on the interval [-2.5, 5].

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justaguide | College Teacher | (Level 2) Distinguished Educator

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We have to find the minimum g(t)=t^3-25t+6 on the interval [-2.5, 5].

First, we find the derivative of g(t). That is equated to 0 to solve for x. The value of the second derivative is positive where the function forms a minimum.

g(t)=t^3-25t+6

g'(t) = 3t^2 - 25

3t^2 - 25 = 0

=> t^2 = 25/3

=> t = -sqrt (25/3) and +sqrt (25/3)

For these + sqrt(25/3) lies in the given interval.

g''(t) for t = sqrt(25/3) is positive indicating that we have a minimum.

The minimum value is [sqrt (25/3)]^3 - 25*sqrt(25/3) + 6

=> (25/3)*(sqrt(25/3) - 25*sqrt(25/3) + 6

=> sqrt(25/3)[25/3 - 25] + 6

=> (-50/3)*sqrt(25/3) + 6

=> -250/3*sqrt 3 + 6

The required minimum of g(t)= t^3-25t+6 on the interval [-2.5, 5] is -250/(3*sqrt 3) + 6

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