posted by help!! on .
find the absolute minimum and maximum values of f(x)=x^(1/3) (3+x) on the interval [-2,1]
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.
do not get it :(
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