SOLUTION: Find the absolute extrema of the function on the closed interval.
y = x^2 − 18 ln x, [1, 6]
minimum (x, y) =
maximum (x, y) =
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-> SOLUTION: Find the absolute extrema of the function on the closed interval.
y = x^2 − 18 ln x, [1, 6]
minimum (x, y) =
maximum (x, y) =
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Question 1187611: Find the absolute extrema of the function on the closed interval.
y = x^2 − 18 ln x, [1, 6]
minimum (x, y) =
maximum (x, y) = Answer by Boreal(15235) (Show Source):
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The first derivative is y'=2x-(18/x)
set that equal to 0 and 2x=(18/x) and 2x^2=18 and x^2=9 and x=+/-3. Only +3 is in the interval.
So the critical values are x=1, 3, 6
f(1)=1, since ln 1=0
f(3)=9-18 ln 3=-10.78 This is the minimum
f(6)=36-18ln 6=3.748 This is the maximum
Graph of the function is
