Q. 85 - 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 $J$.

$f\left(x\right)=\frac{\sin x}{1-\cos x},I=(0,\pi ],J=(0,2\pi )$.

Step-by-Step Solution

Answer

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

On $J,f$ has no global minimum and no global maximum, since $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\infty$ and $\underset{x\to 2{\pi }^{-}}{\lim }f\left(x\right)=-\infty$.

Step 1 . Given information

$f\left(x\right)=\frac{\sin x}{1-\cos x},I=(0,\pi ],J=(0,2\pi )$.

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

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

This means a infinite number of critical points are there.

Step 3 . Again,

$\underset{x\to 0^{+}}{\lim }f\left(x\right)=\underset{x\to 0^{+}}{\lim }\frac{\sin x}{1-\cos x}$  [ in the form of $\frac{\infty }{\infty }$]

                $=\underset{x\to 0^{+}}{\lim }\frac{\cos x}{-(-\sin x)}$ [ Using L'Hopital's rule]

                $$\begin{gathered} =\underset{x\to 0^{+}}{\lim }\frac{\cos x}{\sin x} \\ =\underset{x\to 0^{+}}{\lim }\cot x \\ =\infty \end{gathered}$$

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

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

Step 5 . Therefore, on I , f has no global maximum and a global minimum at x = π , since lim x → 0 + f ( x ) = ∞ .

Again,

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

                  $=\underset{x\to 2{\pi }^{-}}{\lim }\frac{\cos x}{-(-\sin x)}$ [ Using L'Hopital's rule]

                   $$\begin{gathered} =\underset{x\to 2{\pi }^{-}}{\lim }\frac{\cos x}{\sin x} \\ =\underset{x\to 2{\pi }^{-}}{\lim }\cot x \\ =-\infty \end{gathered}$$

Step 6 . The graph for the function with limit J = 0 , 2 π is shown below:

Therefore, on $J,f$ has no global minimum and no global maximum, since $\underset{x\to 0^{+}}{\lim }f\left(x\right)=\infty$ and $\underset{x\to 2{\pi }^{-}}{\lim }f\left(x\right)=-\infty$.