How do use the first derivative test to determine the local extrema $f (x) = \frac{x}{x^{2} + 1}$ $f(x) = x / (x^2+1)$ ?

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Answer 1 · 2015-09-29T05:53:51

See the explanation.

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

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

$f'(x)>0$ $AAx in (-1,1)$ function increasing
$f'(x)<0$ $AAx in (-oo,-1)uu(1,oo)$ function decreasing

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

$f'(x)$ changes sign at $x=1$ and $f(x)$ has a maximum value $f_max=f(1)=1/2$

How do use the first derivative test to determine the local extrema f(x) = x / (x^2+1)f(x)=xx2+1f(x) = x / (x^2+1)?