# 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.

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.