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

Question:

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

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

Critical Points:

We can find the critical points of a function when we equal the first derivative to zero or when the first derivative does not exist. Also, we can classify the critical points found in maximum and minimum.

Answer and Explanation:

We have the function

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

Differentiating the function

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

{eq}f'(x)=0 {/eq} when {eq}x= \frac{5}{3} \pi \\ x= \frac{1}{3} \pi \\ x=- \frac{1}{3} \pi \\ x= - \frac{5}{3} \pi \\ x=\frac{7}{3} \pi \\ x=-\frac{7}{3} \pi \\ {/eq}

Critical points at:

{eq}( \frac{5}{3} \pi , - \frac{1}{3} \sqrt{ 3} ) \\ (- \frac{5}{3} \pi , \frac{1}{3} \sqrt{ 3} ) \\ (\frac{1}{3} \pi, \frac{1}{3} \sqrt{ 3}) \\ ( -\frac{1}{3} \pi, - \frac{1}{3} \sqrt{ 3}) \\ (\frac{7}{3} \pi , \frac{1}{3} \sqrt{ 3} ) \\ (-\frac{7}{3} \pi , - \frac{1}{3} \sqrt{ 3} ) \\ {/eq}

Draw the function:


Learn more about this topic:

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Finding Critical Points in Calculus: Function & Graph

from CAHSEE Math Exam: Tutoring Solution

Chapter 8 / Lesson 9
225K

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