f(x) = \frac{k(1 + c^2x^3)}{(1 + x)^3} Find the critical number(s) for this function.

Question:

{eq}f(x) = \frac{k(1 + c^2x^3)}{(1 + x)^3} {/eq}

Find the critical number(s) for this function.

Critical Points:

The critical points of the function can be in terms of the arbitrary constant, given the equation in terms fo the arbitrary constants. The first - order derivative method is to be also known to find the critical points.

Answer and Explanation:


To fond the critical point, we need to find the derivative of the function and solve it for zero, as follows;

{eq}\displaystyle f'(x)=0\\ \displaystyle \Rightarrow \frac{d}{dx}\left(\:\frac{k\left(1\:+\:c^2x^3\right)}{\left(1\:+\:x\right)^3}\right)=0\\ \displaystyle \Rightarrow k\frac{\frac{d}{dx}\left(1+c^2x^3\right)\left(1+x\right)^3-\frac{d}{dx}\left(\left(1+x\right)^3\right)\left(1+c^2x^3\right)}{\left(\left(1+x\right)^3\right)^2}=0~~~~~~~~~~~~~~~~~~\left [ \because \left(\frac{f}{g}\right)'=\frac{f\:'\cdot g-g'\cdot f}{g^2} \right ]\\ \displaystyle \Rightarrow k\frac{3c^2x^2\left(1+x\right)^3-3\left(1+x\right)^2\left(1+c^2x^3\right)}{\left(\left(1+x\right)^3\right)^2}=0\\ \displaystyle \Rightarrow \frac{3k\left(c^2x^2-1\right)}{\left(1+x\right)^4}=0\\ \displaystyle \Rightarrow x=\pm \frac{1}{c} {/eq}

are the two critical points for x.


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
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