SOLUTION: Find the Absolute maximum and minimum values of f on the given interval. Give exact answers.
f(x)=ln x/x^2 on [1,3]
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Question 980385: Find the Absolute maximum and minimum values of f on the given interval. Give exact answers.
f(x)=ln x/x^2 on [1,3]
Found 2 solutions by Fombitz, htmentor:
Answer by Fombitz(32388) (Show Source): You can put this solution on YOUR website!
find the derivative of f.
The absolute maximum in the interval is the absolute maximum and occurs when
when
and the absolute minimum in the interval occurs at
when
.
.
.
Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
f(x) = ln(x)/x^2 on [1,3]
Since the denominator is always positive, and the minimum value of ln(x) on this interval occurs at x=1: ln(1) = 0
The maximum value is obtained by taking the derivative and setting=0:
df/dx = 0 = 1/x^3 - 2*ln(x)/x^3 -> 2*ln(x) = 1 -> ln(x) = 1/2
exp(ln(x)) = x = exp(1/2)
f(exp(1/2) = ln(exp(1/2)/(exp(1/2))^2 = (1/2)/(exp(1/2))^2 (~0.184)
So minimum value = 0 and maximum value = (1/2)/(exp(1/2))^2 (~0.184)
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