What is the critical f(x)=2x^4-3x^3+x^2 ?

The critical f(x)=2x^4-3x^3+x^2 is x=0,x=(9-sqrt(17))/(16),x=(9+sqrt(17))/(16)

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critical f(x)=2x^4-3x^3+x^2

critical points

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

x = 0, x = \frac{9 - 17}{16}, x = \frac{9 + 17}{16}

Solution steps

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

Find where

f^{'} (x)

is equal to zero or undefined

x = 0, x = \frac{9 - 17}{16}, x = \frac{9 + 17}{16}

Identify critical points not in

f (x)

domain

Domain of

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

All critical points are in domain

x = 0, x = \frac{9 - 17}{16}, x = \frac{9 + 17}{16}

f (x) = - 0.1 x^{2} + 1.4 x + 98.4

f (x) = (x^{7} + 4)^{\frac{1}{5}} - 2

f (x) = \frac{x}{x ( x + 3 )}

f (x) = - \frac{1}{5} sin (\frac{x}{3})

\frac{x}{x - 2}

What is the critical f(x)=2x^4-3x^3+x^2 ?
The critical f(x)=2x^4-3x^3+x^2 is x=0,x=(9-sqrt(17))/(16),x=(9+sqrt(17))/(16)