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 $J$.

$f\left(x\right)=\frac{e^x}{1+x^2},1=[0,\infty ),J=(-\infty ,\infty )$.

Short answer

On $I,f$has no global maximum and a global minimum at $x=0$, since$\underset{x\to \infty }{\lim }f\left(x\right)=\infty$.

On $J,f$has a global minimum at $x=-4$and no global maximum, since $\underset{x\to \infty }{\lim }f\left(x\right)=\infty$.

Step-by-step solution

Step 1. Given information

$f\left(x\right)=\frac{e^x}{1+x^2},1=[0,\infty ),J=(-\infty ,\infty )$.

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

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

Therefore, the function has a critical point at $x=1$.

Again,

$\underset{x\to \infty }{\lim }f\left(x\right)=\underset{x\to \infty }{\lim }\frac{e^x}{1+x^2}$[ in the form of $\frac{\infty }{\infty }$]

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

$=\underset{x\to \infty }{\lim }\frac{e^x}{2}$ [ L'Hopital's rule]

$=\infty$

Step 3. The graph of the function with limit I=0,∞ is shown below:

Step 4.Therefore, on I,fhas no global maximum and a global minimum at  x=0 , since limx→∞f(x)=∞.

Again,

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

Step 5. The graph of the function with limit J=-∞,∞ is shown below:

Therefore, on $J,f$has a global minimum at $x=-4$ and no global maximum, since$\underset{x\to \infty }{\lim }f\left(x\right)=\infty$.