Question

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

Answer 1

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!