# Draw the following function and find its critical points (if any). f(x) = \frac {\sin x}{\cos x-2}

## Question:

Draw the following function and find its critical points (if any).

{eq}f(x) = \frac {\sin x}{\cos x-2} {/eq}

## Trigonometric Function and Its Critical Points

A critical point is a point as the first derivative is zero or undefined, also, the first derivative tells us if the function is increasing or decreasing. Furthermore, a trigonometric function has repetitive behavior, therefore, some trigonometric functions have infinite critical points.

The function is:

{eq}\displaystyle \ f(x) \, = \, \frac {\sin x}{\cos x-2} {/eq}

Its first derivative is:

{eq}\displaystyle \ f'(x) \; = \; {\frac {\cos \left( x \right) }{\cos \left( x \right) -2}}+{\frac { \left( \sin \left( x \right) \right) ^{2}}{ \left( \cos \left( x \right) -2 \right) ^{2}}} \\ \displaystyle \ f'(x) \; = \; {\frac { \left( \cos \left( x \right) \right) ^{2}+ \left( \sin \left( x \right) \right) ^{2}-2\,\cos \left( x \right) }{ \left( \cos \left( x \right) -2 \right) ^{2}}} {/eq}

The critical point at first derivative zero:

{eq}\displaystyle 0={\frac { \left( \cos \left( x \right) \right) ^{2}+ \left( \sin \left( x \right) \right) ^{2}-2\,\cos \left( x \right) }{ \left( \cos \left( x \right) -2 \right) ^{2}}} \; \Leftrightarrow \; x= -\frac{ 7\pi }{ 3 } \; \text{ , } \; x= - \frac{ \pi }{ 3 } \; \text{ , } \; x= \frac{ 5\pi }{ 3 } \; \text{ ... } x= - \frac{ \pi }{ 3 } + 2 \pi \mathbb{N} \\ \; \text{ & } \; \\ \displaystyle x= -\frac{ 4\pi }{ 3 } \; \text{ , } \; x= \frac{ 2 \pi }{ 3 } \; \text{ , } \; x= \frac{ 8\pi }{ 3 } \; \text{ ... } x= - \frac{ 4\pi }{ 3 } + 2 \pi \mathbb{N} \\ {/eq}

Few critical points are:

{eq}\displaystyle \left(- \frac{ 7\pi }{ 3 } , f \left( - \frac{ 7\pi }{ 3 } \right) \right) \; \Rightarrow \; \left( - \frac{ 7\pi }{ 3 } , \frac{ \sqrt{ 3 } }{ 3 } \right) \\ \displaystyle \left( - \frac{ 4 \pi }{ 3 } , f \left( - \frac{ 4\pi }{ 3 } \right) \right) \; \Rightarrow \; \left( - \frac{ 4 \pi }{ 3 } , -\frac{ \sqrt{ 3 } }{ 5 } \right) \\\displaystyle \left( - \frac{ \pi }{ 3 } , f \left( - \frac{ \pi }{ 3 } \right) \right) \; \Rightarrow \; \left( - \frac{ \pi }{ 3 } , \frac{ \sqrt{ 3 } }{ 3 } \right) \\ \displaystyle \left( \frac{ 2\pi }{ 3 } , f \left( \frac{ 2\pi }{ 3 } \right) \right) \; \Rightarrow \; \left( \frac{ 2\pi }{ 3 } , -\frac{ \sqrt{ 3 } }{ 5 } \right) \\ {/eq}

Graph

Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
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