What is the critical 12x^5+15x^4-240x^3+6 ?

The critical 12x^5+15x^4-240x^3+6 is x=-4,x=0,x=3

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critical 12x^5+15x^4-240x^3+6

critical points

12 x^{5} + 15 x^{4} - 240 x^{3} + 6

x = - 4, x = 0, x = 3

Solution steps

12 x^{5} + 15 x^{4} - 240 x^{3} + 6

Find where

f^{'} (x)

is equal to zero or undefined

x = - 4, x = 0, x = 3

Identify critical points not in

f (x)

domain

Domain of

12 x^{5} + 15 x^{4} - 240 x^{3} + 6 : - \infty < x < \infty

All critical points are in domain

x = - 4, x = 0, x = 3

f (x) = 3 sec (\frac{2}{3} x)

- cos (2 x)

y = 2 x + 7

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

\frac{5 ( x + 5 )}{x}

What is the critical 12x^5+15x^4-240x^3+6 ?
The critical 12x^5+15x^4-240x^3+6 is x=-4,x=0,x=3