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]

Algebra ->  Absolute-value -> 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]      Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
find the derivative of f.
df%2Fdx=%28x%5E2%2A%281%2Fx%29-ln%28x%29%2A2x%29%2Fx%5E4
df%2Fdx=%28x-2xln%28x%29%29%2Fx%5E4
df%2Fdx=%281-2ln%28x%29%29%2Fx%5E3
The absolute maximum in the interval is the absolute maximum and occurs when
1-2ln%28x%29=0
1=2ln%28x%29
ln%28x%29=1%2F2
x=e%5E%281%2F2%29
when
y=ln%28e%5E%281%2F2%29%29%2Fe
y=1%2F%282e%29
and the absolute minimum in the interval occurs at x=1
when
y=ln%281%29%2F1%5E2
y=0
.
.
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Answer by htmentor(1343) About Me  (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)