Jtdylktuy

Determine the intervals where the function is increasing and where it is decreasing. (Enter your answers using interval notation.)

$$f(x)=frac{ln(x)}{x}$$

Find intervals of concavity for the graph of the function

$$f(x)=frac{ln(x)}{x}$$

I have already found the first and second derivative but I am confused on how to solve for it

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

$$f ''(x) = frac{2ln(x)-3}{x^3}$$

So let's first begin with finding out the intervals where the function is increasing and decreasing.

To do this, we would find the critical numbers of the first derivative.
(That is, what values of x, set f'(x) = 0? Solving this, we get:

x= e and x =0

Now that you have your critical numbers, you can form your intervals. This would be: (negative infinity to 0, 0 to e and e to infinity)

You do the same thing for the second derivative to find intervals for concavity

Hint

You have got $$f'(x)=dfrac{1-ln x}{x^2}.$$ If you solve $1-ln x=0$ you get $x=e.$ Note that $x^2$ is positive on $(0,infty)$ which is the domain of $f.$ So, you need to study the sign of $f$ on $(0,e)$ and $(e,infty).$ You get

$$begin{array}{ccc} & (0,e) & (e,infty)\ mathrm{sign}(f') & quad+ & -end{array}$$ Thus, $f$ is increasing on $(0,e)$ and decreasing on $(e,infty).$

Proceed in a similar way with the second derivative. (Note that $x^3$ is positive on $(0,infty)$ and so you only need to study the sign of the numerator.)

Intervals of concavity:

$$f''(x)<0 iff ln(x)<3/2 iff ln(x)<ln(e^{3/2}) iff x<e^{3/2}approx 4.48.$$

(the last equivalence being due to the increasing property of $ln$ function).

Thus $begin{cases}0<x<e^{3/2} &Longrightarrow & f text{concave,}\x>e^{3/2} &Longrightarrow & f text{convex.}end{cases}$

One can see/check on the graphics below that point $A$ with abscissa $e^{3/2}$ separates the curve into its concave and convex parts.