Posted by **help!!** on Sunday, April 28, 2013 at 3:42am.

find the absolute minimum and maximum values of f(x)=x^(1/3) (3+x) on the interval [-2,1]

- calculus -
**Steve**, Sunday, April 28, 2013 at 6:06am
f' = (4x+3)/(3x^(2/3))

This is undefined at x=0, where f(0) = 0, a local min.

f'=0 at x = -3/4 where there is a local max. f(-3/4) = -(9/4) ∛(3/4)

check the value of f at the end points for other max/min values.

- calculus -
**sonia**, Sunday, April 28, 2013 at 2:54pm
do not get it :(

- calculus -
**Steve**, Sunday, April 28, 2013 at 3:46pm
what don't you get?

max/min occurs where f' = 0, as long as f" is not also zero

where f' is undefined, check to see whether f is defined there. If so, it will be some kind of cusp, so a max or min. Or it will have a vertical tangent.

As it turns out, I made a mistake at x=0; there is no local min there.

And, the extremum at x = -3/4 is a min, not a max.

Oddly, this error was caused by my trust in the graphs at wolframalpha.com, which showed it wrong.

To see the real graph, visit

http://rechneronline.de/function-graphs/

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