Q. 84 - Calculus

Question

Now that we know L’Hopital’s rule, we can apply it to solve more sophisticated global optimization problems. Consider domains, limits, derivatives, and values to determine the global extrema of each function $f$ in Exercises 81–86 on the given intervals $I$ and $J$.

$f\left(x\right)=\frac{\ln x}{\ln 2x},I=[0,1],J=[1,\infty )$.

Step-by-Step Solution

Answer

On $I,f$ has neither a global maximum nor a global minimum. 

On $J,f$ has neither a global maximum nor a global minimum. 

Step 1 . Given information

$f\left(x\right)=\frac{\ln x}{\ln 2x},I=[0,1],J=[1,\infty )$.

Step 2 . Now, for critical point, f ' x = 0 .

$$\begin{gathered} \frac{d}{dx}\left(\frac{\ln x}{\ln 2x}\right)=0 \\ \frac{\ln \left(2x\right)·\frac{1}{x}-\ln x·\frac{1}{2x}·2}{(\ln 2x{)}^2}=0 \\ \frac{1}{x}\ln \left(2x\right)-\frac{1}{x}\ln x=0 \\ \ln 2x-\ln x=0 \\ \ln \frac{2x}{x}=0 \\ \ln 2=0 \end{gathered}$$

This is a false statement. Therefore, the function has no critical point.

Step 3 . Again,

$\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\frac{\ln x}{\ln 2x}$ [ in the form of $\frac{\infty }{\infty }$]

              $=\underset{x\to 0}{\lim }\frac{\frac{1}{x}}{\frac{1}{2x}·2}$ [Using L'Hopital's rule]

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

Step 4 . The graph of the function with limit I = 0 , 1 is shown below:

Step 5 . Therefore, on I , f has neither a global maximum nor a global minimum.

Again,

$\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{\ln x}{\ln 2x}$  [ in the form of $\frac{\infty }{\infty }$]

                $=\underset{x\to \infty }{\lim }\frac{\frac{1}{x}}{\frac{1}{2x}·2}$ [ Using L'Hopital's rule]

                 $$\begin{gathered} =\underset{x\to \infty }{\lim }\frac{\frac{1}{x}}{\frac{1}{x}} \\ =\underset{x\to \infty }{\lim }1 \\ =1 \end{gathered}$$

Step 6 . The graph for the function with limit J = 1 , ∞ is shown below:

Therefore, on $J,f$ has neither a global maximum nor a global minimum.