# How do you find the critical points of f when f(x)= x^4/(5(x-5)^2)?

May 25, 2015

Critical points are x=0, x= 10

First get f'(x)= $\frac{1}{5} \left[\frac{4 {x}^{3} {\left(x - 5\right)}^{2} - {x}^{4} 2 \left(x - 5\right)}{x - 5} ^ 4\right]$

$\frac{2 {x}^{3} \left(x - 5\right)}{5} \left[\frac{2 \left(x - 5\right) - x}{x - 5} ^ 4\right]$

$\frac{2 {x}^{3}}{5} \frac{x - 10}{x - 5} ^ 3$

Solving for f(x)=0, it is x=0, x=10

At x=5, f'(x) does not exist, hence x=5 is also a critical point.