For #f(x)=y = x^(2/3)(x^2-16) #, the critical points are #0, 2, -2#

**Solution**

#c# is a critcal point for #f# if #c# is in the domain of #f# and either #f'(c)=0# or #f'(c)# does not exist.

**Find #f'(x)#**

We have choices for finding #f'(x)#.

We could leave the function as written and use the product rule, or we could distribute the #x^(2/3)# and avoid the product rule.

**Using the product rule** #("I use:"(FS)'=F'S+FS')#

#f'(x)=2/3x^(-1/3) (x^2-16) + x^(2/3)(2x)#

#=(2(x^2-16))/(3root(3)x)+(2x*x^(2/3))/(1)#

#=(2(x^2-16))/(3root(3)x)+(6x^2)/(3root(3)x)=(8x^2-32)/(3root(3)x)#

**.Re-writing #f# before differentiating**

#f(x) = x^(8/3)-16x^(2/3) #

#f'(x) = 8/3 x^(5/3)-16*2/3 x^(-1/3) =(8 x^(5/3))/3- 32/(3 root(3) x)#

#=(8 x^(5/3)*x^(1/3))/(3 root(3)x)- 32/(3 root(3) x)= (8x^2-32)/(3root(3)x)#

.

Using either method, we get #f'(x)= (8x^2-32)/(3root(3)x) #

**Critical points:**

#c# is a critcal point for #f# if #c# is in the domain of #f# and either #f'(c)=0# or #f'(c)# does not exist.

#f(x)=y = x^(2/3)(x^2-16) #

For this function, the domain is #(-oo, oo)#

So the critical points for this #f# will be the zeros of #f'#. and all zeros of the denominator of #f'#

#f'(x)=(8x^2-32)/(3root(3)x)=0# when #8(x^2-4)=0#

at #x=+-2#,

and #f'(x)# fails to exist at #x=0#

Because these are all in the domain of #f#, these are all critical points for this #f#.