Question

How do you find the monotonicity, extrema, concavity, and inflection points of \[f(x) = \dfrac{{\ln x}}{{\sqrt x }}?\]

Answer 1

Hint: First, find the domain of the function then find its range by putting the boundaries in the of domain in the function to get the extrema and then with help of first derivative find the monotonicity of the function and at last, find concavity and inflection point with help of the second derivative of the function.

Complete step by step solution:
In order to find the monotonicity, extrema, concavity, and inflection points of \[f(x) = \dfrac{{\ln x}}{{\sqrt x }}\], we will first find domain of the function as follows
Since log function has positive real numbers as its domain and root too has same domain, so the domain of the function will be given as $ x \in (0,\;\infty ) $
Now, finding range of the function
 $
  \mathop {\lim }\limits_{x \to {0^ + }} f(x) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{{\ln x}}{{\sqrt x }} = - \infty \\
  \mathop {\lim }\limits_{x \to \infty } f(x) = \mathop {\lim }\limits_{x \to \infty } \dfrac{{\ln x}}{{\sqrt x }} = 0 \\
  $
So the function has a vertical asymptote as $ x = 0 $ and $ y = 0 $ as its horizontal asymptote.
Now, we will calculate first and second derivative of the function in order to find the monotonicity, extrema, concavity, and inflection point
Finding the first derivative,
\[
  f(x) = \dfrac{{\ln x}}{{\sqrt x }} \\
  f'(x) = \dfrac{{\dfrac{{\sqrt x }}{x} - \dfrac{{\ln x}}{{2\sqrt x }}}}{x} = \dfrac{{\sqrt x }}{{{x^2}}} - \dfrac{{\ln x}}{{2x\sqrt x }} = \dfrac{{2 - \ln x}}{{2x\sqrt x }} \\
 \]
We can see that $ f'(x) > 0\;{\text{if}}\;2 - \ln x > 0 \Rightarrow x < {e^2} $ that means the function is increasing from $ 0\;{\text{to}}\;{e^2} $ and decreasing further from $ {e^2} $ up to infinity. It also means, function has a local extrema at $ x = {e^2} $
Finding the second derivative
\[
  f(x) = \dfrac{{\ln x}}{{\sqrt x }} \\
  f''(x) = \dfrac{{\left( { - \dfrac{{2x\sqrt x }}{x}} \right) - \left( {(2 - \ln x)3\sqrt x } \right)}}{{4{x^3}}} = \dfrac{{ - 2\sqrt x - 6\sqrt x + 3\ln x\sqrt x }}{{4{x^3}}} = \dfrac{{3\ln x - 8}}{{4{x^2}\sqrt x }} \\
 \]

We can see that $ f''(x) > 0\;{\text{if}}\;3\ln x - 8 > 0 \Rightarrow x > {e^{\dfrac{8}{3}}} $ that means the function is concave down in interval \[\left( {0,\;{e^{\dfrac{8}{3}}}} \right)\] and concave up in $ \left( {{e^{\dfrac{8}{3}}},\;\infty } \right) $ also inflection point will be $ x = {e^{\dfrac{8}{3}}} $

Note: It is not necessary that a function will have its minima at the left boundary of the domain and maxima or extrema at the right boundary of the domain, functions may have their maximum value or local maximum in between the domain. So always check for monotonicity of function when finding its minima or maxima.