What is the critical 8x^{1/3}+x^{4/3} ?

The critical 8x^{1/3}+x^{4/3} is x=-2,x=0

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critical 8x^{1/3}+x^{4/3}

critical points

8 x^{\frac{1}{3}} + x^{\frac{4}{3}}

x = - 2, x = 0

Solution steps

8 x^{\frac{1}{3}} + x^{\frac{4}{3}}

Find where

f^{'} (x)

is equal to zero or undefined

x = - 2, x = 0

Identify critical points not in

f (x)

domain

Domain of

8 x^{\frac{1}{3}} + x^{\frac{4}{3}} : - \infty < x < \infty

All critical points are in domain

x = - 2, x = 0

f (x) = \frac{x ^{2} - 7}{16 - x ^{2}}

f (x) = 2 x^{2} - 5 x + 1

- 9 - x^{2}

- 2 (x - 3)^{2} + 7

f (x) = \frac{x ^{5} + 10}{3}

What is the critical 8x^{1/3}+x^{4/3} ?
The critical 8x^{1/3}+x^{4/3} is x=-2,x=0