How do you find the critical numbers of $x^4-2x^3-3x^2-5$ ?

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Answer 1 · 2018-06-04T05:58:12

Please see the explanation below

Let

$f(x)=x^4-2x^3-3x^2-5$

The first derivative is

$f'(x)=4x^3-6x^2-6x$

$=2x(2x^2-3x-3)$

The critical points are when $f'(x)=0$

$2x(2x^2-3x-3)=0$

$=>$ , ${(x=0),(2x^2-3x-3=0):}$

$=>$ , ${(x=0),(x=(3+sqrt(33))/(4)=2.186),(x=(3-sqrt(33))/4=-0.686):}$

The critical points are $(0,-5)$ , $(2.186, -17.393)$ , and $(-0.686,-5.545)$

graph{x^4-2x^3-3x^2-5 [-24.14, 27.18, -19.11, 6.56]}

How do you find the critical numbers of x^4-2x^3-3x^2-5x^4-2x^3-3x^2-5?