# What are the values and types of the critical points, if any, of f(x)=x^4/(6x^2+x)?

May 11, 2018

x=-1/4 is the minimum of the function.

#### Explanation:

f(x)=(x⁴)/(6x²+x)
We search $f ' \left(x\right) = 0$
f'(x)=(4x³(6x²+x)-(x⁴(12x+1)))/((6x²+x)²)
(4x³(6x²+x)-(x⁴(12x+1)))/((6x²+x)²)=0
4x³(6x²+x)-(x⁴(12x+1))=0
24x^5+4x⁴-12x^5-x^4=0
$12 {x}^{5} + 3 {x}^{4} = 0$
$4 {x}^{5} + {x}^{4} = 0$
$4 x + 1 = 0$
$x = - \frac{1}{4}$
Now let study lim_(x to ±oo) f(x)
=lim_(x to ±oo) (x⁴)/(6x²+x)
=lim_(x to ±oo) (x⁴)/(6x²)
=lim_(x to ±oo) (x²)/6
$= + \infty$
Now we can see that f(x) is positive in ±oo#, so $x = - \frac{1}{4}$ is the minimum of the function.