What is the critical x^4-7x^3 ?

The critical x^4-7x^3 is x=0,x= 21/4

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critical x^4-7x^3

critical points

x^{4} - 7 x^{3}

x = 0, x = \frac{21}{4}

Solution steps

x^{4} - 7 x^{3}

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = \frac{21}{4}

Identify critical points not in

f (x)

domain

Domain of

x^{4} - 7 x^{3} : - \infty < x < \infty

All critical points are in domain

x = 0, x = \frac{21}{4}

f (x) = x^{4} + y^{4}

f (x) = x^{3} + 6 x^{2} - 8 y + 25 = 13

f (x) = e^{4 x} (80 x^{2} + 5)

x y + 5 x - 5

f (x, y) = 4 x - 9 y + 2 x y - x^{2} + y^{3}