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Calculus/critical points and abs max and min
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Question
find all critical points, and the absolute maximum and
absolute minimum, of the function f(x)= 5x^(2/3)-x^(5/3) on
the interval [-1,4]
The derivative of the function is
f'(x) = (10/3)/x^(1/3) - (5/3)x^(2/3).
This is the same as f'(x) = 10/[3x^(1/3)] - 5x^(2/3)/3.
Factor out as (5/[3x^(1/3)]) and get
f'(x) = (5/[3x^(1/3)])(2 - 5x).
This is the same as f'(x) = 5(2-5x)/[3x^(1/3)].
To find f"(x), realize that f'(x) = g(x)/h(x) where g(x) = 5(2-5x) and h(x) = 3x^(1/3).
To find the critical points, set f'(x) equal to 0. Where they are 0, find the value of f(x). Also find f(-1) and f(4) since these may be the max and/or min of the function on that interval.
At the points where f'(x) is 0, the sign of f"(x) is of interest.
A positive f"(x) at that point means the point is a local minimum and a negative f"(x) at that point means the point is a local maximum.
Where f"(x) = 0 implies there is an inflection point.
Remember that since f"(x) = (h(x)g'(x) = g(x)h'(x))/h²(x),
just take g'(x) = 5(-5) = -25 and h'(x) = x^(-2/3).
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