How do you find the points where the graph of the function #f(x) = (x^3 + 2) / (x^(1/3))# has horizontal tangents and what is the equation?

First, simplify #f(x)# by splitting up the fraction.

#f(x)=x^3/x^(1/3)+2/x^(1/3)#

#f(x)=x^(8/3)+2x^(-1/3)#

In order to find the horizontal tangents of a function, we must find the times when the derivative of the function equals #0#.

To differentiate #f(x)#, use the power rule.

#f'(x)=8/3x^(5/3)-2/3x^(-4/3)#

It will be easier to find when this equals #0# if we put it into fractional form. First, simplify by putting the #3# in the denominator of a combined fraction.

#f'(x)=(8x^(5/3)-2x^(-4/3))/3#

Now, multiply the fraction by #x^(4/3)/x^(4/3)#.

#f'(x)=(8x^3-2)/(3x^(4/3))#

We can now set this equal to #0# to find the original function's horizontal tangents.

#(8x^3-2)/(3x^(4/3))=0#

#8x^3-2=0#

#x^3=1/4#

#x=4^(-1/3)#

However, the line of the horizontal tangent will be in the form #y=ul(" ")#. Thus, we must find the function value at #x=4^(-1/3)#.

#f(4^(-1/3))=((4^(-1/3))^3+2)/((4^(-1/3))^(1/3))=(1/4+2)(4^(1/9))=9/4(4^(1/9))=9/(4^(8/9))#

The horizontal tangent is the line #y=9/4^(8/9)#.

Graphed are the function and its tangent line:

graph{((x^3 + 2) / (x^(1/3))-y)(y-9/4^(8/9)-0.0001x)=0 [-19.65, 20.9, -8.3, 12]}