How do you find the critical values for $f(x)=x^(2/3)+x^(-1/3)$ ?

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Answer 1 · 2015-10-26T18:43:32

Find values in the domain of $f$ at which $f'(x)$ does not exist or $f'(x)=0$ . See the explanation.

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

$"Dom"(f) = RR - {0}$

$f'(x) = 2/3x^(-1/3) - 1/3x^(-4/3)$

$=1/3 x^(-4/3) (2x-1)$

$= (2x-1)/(3x^(4/3))$

$f'(x)$ does not exist at $x=0$ , which is not in $"Dom"(f)$ .

$f'(x) = 0$ at $x=1/2$ which is in $"Dom"(f)$ .

The only critical number for $f$ is $1/2$ .

How do you find the critical values for f(x)=x^(2/3)+x^(-1/3)f(x)=x^(2/3)+x^(-1/3)?