What is the critical y=x^{4/7}(x^2-6) ?

The critical y=x^{4/7}(x^2-6) is x=-2/(3^{1/2)},x=0,x= 2/(3^{1/2)}

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critical y=x^{4/7}(x^2-6)

critical points

y = x^{\frac{4}{7}} (x^{2} - 6)

x = - \frac{2}{3 ^{\frac{1}{2}}}, x = 0, x = \frac{2}{3 ^{\frac{1}{2}}}

Solution steps

x^{\frac{4}{7}} (x^{2} - 6)

Find where

f^{'} (x)

is equal to zero or undefined

x = - \frac{2}{3 ^{\frac{1}{2}}}, x = \frac{2}{3 ^{\frac{1}{2}}}, x = 0

Identify critical points not in

f (x)

domain

Domain of

x^{\frac{4}{7}} (x^{2} - 6) : - \infty < x < \infty

All critical points are in domain

x = - \frac{2}{3 ^{\frac{1}{2}}}, x = 0, x = \frac{2}{3 ^{\frac{1}{2}}}

f (x) = \frac{x ^{3}}{3} - 3 x^{2} + 8 x - 4

f (x) = (x^{2} - 1) y

f (x) = - x^{3} + 4 x^{2} + 5 x

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

\frac{x ^{7} - 4 x ^{6} - 16 x ^{5} + 64 x ^{4}}{x}

What is the critical y=x^{4/7}(x^2-6) ?
The critical y=x^{4/7}(x^2-6) is x=-2/(3^{1/2)},x=0,x= 2/(3^{1/2)}