What are the local extrema, if any, of #f(x) =(lnx-1)^2 / x#?

Question

What are the local extrema, if any, of #f(x) =(lnx-1)^2  / x#?

Charles Anctil · Accepted Answer

#(e^3, 4e^-3)# Maximum Point

#(e, 0)# Minimum Point

Paisley Johnson · Answer

undefined To find the local extrema of $ f(x) = \frac{{(\ln x - 1)^2}}{{x}} $, we need to take the derivative of $ f(x) $ and set it equal to zero.

First, let's find the derivative of $ f(x) $ using the quotient rule:

$$ f'(x) = \frac{{x(2(\ln x - 1)\frac{1}{x} - (\ln x - 1)^2 \cdot \frac{1}{x^2}) - (\ln x - 1)^2}}{{x^2}} $$

$$ f'(x) = \frac{{2(\ln x - 1) - (\ln x - 1)^2}}{{x^2}} $$

Now, we set $ f'(x) $ equal to zero and solve for $ x $:

$$ 2(\ln x - 1) - (\ln x - 1)^2 = 0 $$

$$ 2\ln x - 2 - (\ln^2 x - 2\ln x + 1) = 0 $$

$$ 2\ln x - 2 - \ln^2 x + 2\ln x - 1 = 0 $$

$$ \ln^2 x - 4\ln x + 3 = 0 $$

$$ (\ln x - 1)(\ln x - 3) = 0 $$

$$ \ln x = 1 \quad \text{or} \quad \ln x = 3 $$

$$ x = e \quad \text{or} \quad x = e^3 $$

To determine whether these are local extrema, we can use the second derivative test.

$$ f''(x) = \frac{{d}}{{dx}}\left(\frac{{2(\ln x - 1) - (\ln x - 1)^2}}{{x^2}}\right) $$

$$ f''(x) = \frac{{2 - 2(\ln x - 1) \cdot \frac{1}{x} - 2(\ln x - 1) \cdot \frac{1}{x} + 2(\ln x - 1)^2 \cdot \frac{1}{x^2} - 2(\ln x - 1)^2 \cdot \frac{2}{x^3}}}{{x^2}} $$

$$ f''(x) = \frac{{2 - 2\frac{1}{x} - 2\frac{1}{x} + 2(\ln x - 1) - 4(\ln x - 1) \cdot \frac{1}{x}}}{{x^2}} $$

$$ f''(x) = \frac{{2 - 4\frac{1}{x} + 2(\ln x - 1) - 4(\ln x - 1) \cdot \frac{1}{x}}}{{x^2}} $$

$$ f''(x) = \frac{{2 - 4\frac{1}{x} + 2\ln x - 2 - 4\ln x + 4}}{{x^2}} $$

$$ f''(x) = \frac{{2\ln x - 2 - 4\frac{1}{x}}}{{x^2}} $$

At $ x = e $:

$$ f''(e) = \frac{{2\ln e - 2 - 4\frac{1}{e}}}{{e^2}} = 0 $$

At $ x = e^3 $:

$$ f''(e^3) = \frac{{2\ln(e^3) - 2 - 4\frac{1}{{e^3}}}}{{(e^3)^2}} = \frac{{6 - 2 - 4/e^3}}{{e^6}} $$

$$ f''(e^3) = \frac{{4 - 4/e^3}}{{e^6}} > 0 $$

Therefore, $ x = e $ corresponds to a local minimum, and $ x = e^3 $ corresponds to a local maximum for the function $ f(x) $.