What are the global and local extrema of $f (x) = x^{3} - x^{2} - x$ $f(x)=x^3-x^2-x$ ?

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What are the global and local extrema of $f (x) = x^{3} - x^{2} - x$ $f(x)=x^3-x^2-x$ ?

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Answer 1 · 2016-08-31T07:01:19

There is a maxima at $x = - \frac{1}{3}$ $x=-1/3$
There is a minima at $x = 1$ $x=1$

Explanation:

Given -

$y = x^{3} - x^{2} - x$ $y=x^3-x^2-x$
$\frac{d y}{d x} = 3 x^{2} - 2 x - 1$ $dy/dx=3x^2-2x-1$
$\frac{d^{2} y}{{d x}^{2}} = 6 x - 2$ $(d^2y)/(dx^2)=6x-2$

Set the first derivative equal to zero

$\frac{d y}{d x} = 0 \Rightarrow 3 x^{2} - 2 x - 1 = 0$ $dy/dx=0 =>3x^2-2x-1=0$

$3 x^{2} - 3 x + x - 1 = 0$ $3x^2-3x+x-1=0$

$3 x (x - 1) + 1 (x - 1) = 0$ $3x(x-1)+1(x-1)=0$
$(3 x + 1) (x - 1) = 0$ $(3x+1)(x-1)=0$
$3 x = - 1$ $3x=-1$
$x = - \frac{1}{3}$ $x=-1/3$
$x - 1 = 0$ $x-1=0$
$x = 1$ $x=1$

At $x = - \frac{1}{3}$ $x=-1/3$

$\frac{d^{2} y}{{d x}^{2}} = 6 (- \frac{1}{3}) - 2 = - 2 - 2 = - 4 < 0$ $(d^2y)/(dx^2)=6(-1/3)-2=-2-2=-4<0$

There is a maxima at $x = - \frac{1}{3}$ $x=-1/3$

$\frac{d^{2} y}{{d x}^{2}} = 6 (1) - 2 = 6 - 2 = 4 > 0$ $(d^2y)/(dx^2)=6(1)-2=6-2=4>0$

There is a minima at $x = 1$ $x=1$

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What are the global and local extrema of $f (x) = x^{3} - x^{2} - x$ $f(x)=x^3-x^2-x$ ?

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Explanation:

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What are the global and local extrema of $f (x) = x^{3} - x^{2} - x$ $f(x)=x^3-x^2-x$ ?