How do use the first derivative test to determine the local extrema $f (x) = x^{3} - x^{2} - x + 3$ $f(x) = x^3 - x^2 - x + 3$ ?

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Answer 1 · 2015-09-26T01:31:57

See the explanation.

$f'=3x^2-2x-1$
$f'=0 <=> 3x^2-2x-1=0$
$3x^2-3x+x-1=0$
$3x(x-1)+x-1=0$
$(x-1)(3x+1)=0$
$x=1 vv x=-1/3$

$a=3 >0$ function is concave.

graph{3x^2-2x-1 [-10, 10, -5, 5]}

$AAx in (-oo,-1/3)uu(1,+oo) f'(x)>0$ function is increasing

$AAx in (-1/3,1) f'(x)<0$ function is decreasing

$f'(x)$ changes sign in $x=-1/3$ and $f(x)$ has a maximim value $f_max=f(-1/3)=86/27$

$f'(x)$ changes sign in $x=1$ and $f(x)$ has a minimum value $f_min=f(1)=2$

How do use the first derivative test to determine the local extrema f(x) = x^3 - x^2 - x + 3f(x)=x3−x2−x+3f(x) = x^3 - x^2 - x + 3?