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

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

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critical f(x)=x^{(7/2)}-7x^2

critical points

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

x = 0, x = 2^{\frac{4}{3}}

Solution steps

x^{\frac{7}{2}} - 7 x^{2}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = 2^{\frac{4}{3}}

Identify critical points not in

f (x)

domain

Domain of

x^{\frac{7}{2}} - 7 x^{2} : x \geq 0

All critical points are in domain

x = 0, x = 2^{\frac{4}{3}}

2 y = x - 3, (- 1, 2)

f (x) = \frac{3 x}{x ^{2} - 25}

- \frac{1}{2} x^{4} - 6 x^{3} - 72 x^{2}

f (x) = 3 x - 14

\frac{a - \frac{2 a - 1}{a}}{\frac{1 - a}{3 a}}

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