What is the critical x^4+x^3-3x^2+1 ?

The critical x^4+x^3-3x^2+1 is x=(-3-sqrt(105))/8 ,x=0,x=(-3+sqrt(105))/8

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critical x^4+x^3-3x^2+1

critical points

x^{4} + x^{3} - 3 x^{2} + 1

x = \frac{- 3 - 105}{8}, x = 0, x = \frac{- 3 + 105}{8}

Solution steps

x^{4} + x^{3} - 3 x^{2} + 1

Find where

f^{'} (x)

is equal to zero or undefined

x = \frac{- 3 - 105}{8}, x = 0, x = \frac{- 3 + 105}{8}

Identify critical points not in

f (x)

domain

Domain of

x^{4} + x^{3} - 3 x^{2} + 1 : - \infty < x < \infty

All critical points are in domain

x = \frac{- 3 - 105}{8}, x = 0, x = \frac{- 3 + 105}{8}

2 x - 4

f (x) = 3 - x^{2}

f (x) = x^{3} + 1

y - 5 = 6 (x + 1)

- \frac{6}{x ^{2}}

What is the critical x^4+x^3-3x^2+1 ?
The critical x^4+x^3-3x^2+1 is x=(-3-sqrt(105))/8 ,x=0,x=(-3+sqrt(105))/8