What are the local extrema of $f(x)= x^3-x+3/x$ ?

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Answer 1 · 2016-12-06T17:16:15

$x_1= -1$ is a maximum
$x_2= 1$ is a minimum

First find the critical points by equating the first derivative to zero:

$f'(x) = 3x^2-1-3/x^2$

$3x^2-1-3/x^2 = 0$

As $x!=0$ we can multiply by $x^2$

$3x^4-x^2-3=0$

$x^2=frac(1+-sqrt(1+24)) 6$

so $x^2=1$ as the other root is negative, and $x=+-1$

Then we look at the sign of the second derivative:

$f''(x) = 6x+6/x^3$

$f''(-1) = -12 <0$

$f''(1) = 12>0$

so that:

$x_1= -1$ is a maximum
$x_2= 1$ is a minimum

graph{x^3-x+3/x [-20, 20, -10, 10]}

What are the local extrema of f(x)= x^3-x+3/xf(x)= x^3-x+3/x?