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

1 Answer
Guillaume L.
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.
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