How do you find the critical points for $f(x)=x^(8)$ and the local max and min?

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Answer 1 · 2018-01-05T03:48:54

Check below.

$f(x)=x^8$

$D_f=RR$

$f$ is continuous in $RR$

$f'(x)=8x^7$ and

$f'(x)=0 <=> x=0$
$x_0=0$ is a critical point.
There is no other critical point because $f'$ is defined in $RR$ .

If $x<0$ then $f'(x)<0$ so $f$ is decreasing in $(-oo,0]$
If $x>0$ then $f'(x)>0$ so $f$ is increasing in $[0,+oo)$

So for $x$ $in$ $RR$ : $f(x)>=f(0)$

Therefore $f$ has global minimum at $x_0=0$ , $f(0)=0$

How do you find the critical points for f(x)=x^(8)f(x)=x^(8) and the local max and min?