For #f(x)=x^(2/5)(x-10)#, I prefer to rewrite the function before differentiating.

#f(x) = x^(7/5)-10x^(2/5)#

So,

#f'(x) =7/5x^(2/5) - 10*2/5x^(-3/5)#

# = (7x^(2/5))/5 - 4/x^(3/5)#

# = (7x-20)/(5x^(3/5))#

#f'(x) = 0# at #x=20/7# which is in the domain of #f#/

#f'(x)# does not exist at #5x^(3/5) = 0# which happens at #x=0#. Note that #0# is in the domain of the original function, #f#, so it is a critical number for #f#.

The critical numbers for #f# are: #0# and #20/7#.

**Additional Notes**

#f"(x)# changes from positive to negative as we move past #0#, fo #f(0) = 0# is a local maximum value of #f#

#f'(x)# changes from negative to positive as we move past #20/7#, so #f(20/7)# is a local minimum value of #f#.

It was not asked, but here is the graph of the function: #f(x)=x^(2/5)(x-10)#,

graph{y =x^(2/5)(x-10) [-21.1, 36.62, -24.43, 4.43]}