Finding the Min & Max of a function

Question

Question: "for the following functions, determine the locations of relative extrema, either minimum or maximum, and find their values"

 

for k(x)=(ln3x)/(2x^2)

 

The process I use is to first find the derivative --> find the critical #'s --> find 2nd derivative & evaluate at critical numbers --> test critical numbers in original equation to get the value of the min and/or max

 

However, I can't seem to get this problem right no matter how many times I've tried...I think it's because I'm not finding the correct derivatives.

 

When I finally looked up the answer I got max @ e/sqrt3 and no minima found...can someone please tell me if this is correct or if it's not correct what the correct answer is? Or if it's correct what the steps to getting to that answer would be?

 

Thank you SO much!

Answer 1

When my students cannot get the right answer, it is usually because the are forgetting the chain rule.  And in this problem, you need the quotient rule, and every time you take the derivative in the quotient rule it will involve the chain rule  (on only the ln in this problem).

 

Here is what I get:   [(bottom)*(deriv of top) - (top)*(deriv of bottom)]/(bottom)²

 [(2x²)*(1/(3x)*3) - (ln(3x)*(4x)]/(2x²)²

This simplifies to    [2x - 4x ln(3x)]/4x⁴

This further simplifies to  [1-2ln(3x)]/2x³

 

Setting the first derivative = 0 (to find location of mins and/or maxs) get 1 - 2ln(3x) = 0,  so 1 = 2ln(3x), so 1/2 = ln(3x), so e^1/2=3x, and x=√e/3

 

There will be a vertical asymptotote at x=0  (x in denom of original problem)

The domain is x>0  (argument of logarithm is always positive as long as working with only real numbers)

Between 0 and √e/3, the first derivative is positive, and after √e/3, the first deriv is negative.  This means that there is a maximum at x=√e/3.   If you need the point, plug x=√3 back into the original equation and get the maximum at at  [√e/3, 9/(4e)]  or approximately (0.550, 0.828)

 

The critical values are the x-values for all mins and maxs AND ALSO all asymptotes and holes.

 

So the Critical Values are √e/3, and 0.

There is a maximum at √e/3

The value of the maximum is 9/(4e).