What are the values and types of the critical points, if any, of f(x)=x^4/(6x^2+x)f(x)=x46x2+x?

1 Answer
May 11, 2018

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

Explanation:

f(x)=(x⁴)/(6x²+x)
We search f'(x)=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
12x^5+3x^4=0
4x^5+x^4=0
4x+1=0
x=-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
=+oo
Now we can see that f(x) is positive in ±oo, so x=-1/4 is the minimum of the function.
\0/ here's our answer!