How do you find the critical numbers of $f(x)=x^(1/5)-x^(-4/5)$ ?

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Answer 1 · 2017-07-03T18:33:20

The only critical number is $x=-4$

The critical numbers of a function are the points where its derivative equals zero, which means they are the solutions of the equation:

$(df)/dx = 0$

$1/5x^(-4/5)+4/5x^(-9/5) = 0$

$1/5x^(-4/5)(1+4/x) = 0$

As $x^(-4/5) !=0$ for any value of $x$ the only solutions are given by:

$1+4/x = 0$

$4/x = -1$

$x=-4$

How do you find the critical numbers of f(x)=x^(1/5)-x^(-4/5) f(x)=x^(1/5)-x^(-4/5)?