What is the critical 12x^5+15x^4-240x^3+1 ?

The critical 12x^5+15x^4-240x^3+1 is x=-4,x=0,x=3

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critical 12x^5+15x^4-240x^3+1

critical points

12 x^{5} + 15 x^{4} - 240 x^{3} + 1

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

Solution steps

12 x^{5} + 15 x^{4} - 240 x^{3} + 1

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} + 1 : - \infty < x < \infty

All critical points are in domain

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

f (x) = 4 x - 5 y = 24

f (x) = 8 x^{2} - 5 x - 2

(4, - 1), (- 3, 3)

\frac{7 x ^{2} + 490 x}{x - 5}

y = 2 x - 2

What is the critical 12x^5+15x^4-240x^3+1 ?
The critical 12x^5+15x^4-240x^3+1 is x=-4,x=0,x=3