What is the critical f(x)=x^2+x^{1/3} ?

The critical f(x)=x^2+x^{1/3} is x=(-1/6)^{3/5},x=0

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critical f(x)=x^2+x^{1/3}

critical points

f (x) = x^{2} + x^{\frac{1}{3}}

x = (- \frac{1}{6})^{\frac{3}{5}}, x = 0

Solution steps

x^{2} + x^{\frac{1}{3}}

Find where

f^{'} (x)

is equal to zero or undefined

x = (- \frac{1}{6})^{\frac{3}{5}}, x = 0

Identify critical points not in

f (x)

domain

Domain of

x^{2} + x^{\frac{1}{3}} : - \infty < x < \infty

All critical points are in domain

x = (- \frac{1}{6})^{\frac{3}{5}}, x = 0

f (x) = 2 x + 2

2 x - x^{2}

- x^{2} + 10 x - 21

f (x) = x^{4} (x - 2)^{3}

x \cdot ln (x)

What is the critical f(x)=x^2+x^{1/3} ?
The critical f(x)=x^2+x^{1/3} is x=(-1/6)^{3/5},x=0