How do you find the critical points for #y = x^(2/3)(x^2-16) #?

Question

Christopher Agresta · Accepted Answer

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#.

Charles Arocha · Answer

undefined To find the critical points of $ y = x^{2/3}(x^2-16) $, we need to first find the derivative of the function, $ y' $. Then, we set $ y' $ equal to zero and solve for $ x $. Finally, we determine the $ x $-values that make $ y' $ undefined.

First, let's find the derivative of $ y $:
$$ y' = \frac{d}{dx}(x^{2/3}(x^2-16)) $$
$$ y' = \frac{d}{dx}(x^{2/3}) \cdot (x^2-16) + x^{2/3} \cdot \frac{d}{dx}(x^2-16) $$
$$ y' = \frac{2}{3}x^{-1/3}(x^2-16) + x^{2/3} \cdot (2x) $$

Next, set $ y' $ equal to zero and solve for $ x $:
$$ \frac{2}{3}x^{-1/3}(x^2-16) + 2x^{5/3} = 0 $$

Now, solve for $ x $.

After finding $ x $-values, check for points where the derivative is undefined by checking where the original function $ y $ is undefined or its derivative is undefined.