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.
