Q. 81 - Calculus

Question

determine the global extrema of  function $f\left(x\right)=x\ln \left(x\right) , I=\left(0, 1\right), J=\left(0, \infty \right)$

Step-by-Step Solution

Answer

on I, f has a global maximum at x=1 and a global minimum at $x=\frac{1}{e}$

on J, f has no global maximum  and a global minimum at $x=\frac{1}{e}$

Step 1. Given information

The given information is $f\left(x\right)=x\ln \left(x\right) , I=\left(0, 1\right), J=\left(0, \infty \right)$

Step 2. Calculate the function

$$\begin{gathered} f\left(x\right)=x\ln \left(x\right) \\ f\text{'}\left(x\right)=0 \\ \frac{d \left(x\ln \left(x\right)\right)}{dx}=0 \\ x·\frac{1}{x}+\ln \left(x·1\right)=0 \\ \ln \left(x\right)=-1 \\ x=\frac{1}{e} \end{gathered}$$

similarly

$$\begin{gathered} \underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }x\ln \left(x\right) \\ =\underset{x\to 0}{\lim }\frac{\ln \left(x\right)}{{\displaystyle \frac{1}{x}}} \\ =\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{1}{x}}}{-{\displaystyle \frac{1}{x^2}}} \\ =\underset{x\to 0}{\lim }\frac{1}{x}\left(-x^2\right) \\ =\underset{x\to 0}{\lim }\left(-x\right) \\ =0 \end{gathered}$$

Therefore on I, f has a global maximum at x=1 and a global minimum at $x=\frac{1}{e}$

Again 

$$\begin{gathered} \underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }x\ln \left(x\right) \\ =\infty \end{gathered}$$

Therefore on J, f has no global maximum, and a global minimum at $x=\frac{1}{e}$ because $\underset{x\to \infty }{\lim }f\left(x\right)=\infty$

Step 3. The solution

Therefore,

on I, f has a global maximum at x=1 and a global minimum at $x=\frac{1}{e}$

On J, f has no global maximum and a global minimum at $x=\frac{1}{e}$