What are the critical points of $f(x) = 6x^(2/3) + x^(5/3)$ ?

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Answer 1 · 2015-10-27T12:23:43

Thee critical numbers for $f$ are $-12/5$ and $0$ ..

For $f(x) = 6x^(2/3) + x^(5/3)$ , we have

$"Dom"(f)=(-oo,oo)$ and

$f'(x) = 12/3x^(-1/3)+5/3x^(2/3)$

$= 1/3 x^(-1/3) (12+5x)$

$= (12+5x)/(3x^(1/3))$ .

$f'(x)$ is undefined at $x=0$ and
$f'(x) = 0$ at $x=-5/12$ .

Both of these numbers are in $"Dom"(f)$ , so both are critical numbers for $f$ .

Alternative Method

Many students find it more clear to do the algebra differently.

$f'(x) = 12/3x^(-1/3)+5/3x^(2/3)$

$= 12/(3root3x)+(5root3x^2)/3$

$= 12/(3root3x)+(5root3x^2)/3 * root3x/root3x$ .

$= 12/(3root3x)+(5x)/(3root3x)$

$= (12+5x)/(3root3x)$

Now find $0$ and $-12/5$ as above.

What are the critical points of f(x) = 6x^(2/3) + x^(5/3)f(x) = 6x^(2/3) + x^(5/3)?