Find the absolute extrema of the function (x^2+4)^1/3 on the interval [-5,4]

Question

anonymous · Answer

The absolute max and min.

anonymous · Answer

take the derivative, find the critical points, check those and also check the endpoints

anonymous · Answer

although there is really not that much to check
this is an even function so it is smallest at $x=0$ by pure reason, no calculus

anonymous · Answer

The derivative I got is $$1/3(X^2+4)^-2/3 (2X)$$
I'm not sure how to solve to get the critical points.

anonymous · Answer

you have $f(x)=\sqrt[3]{x^2+4}$ and it is pretty obvious that this is largest when $x^2+4$ is largest, and smallest when $x^2+4$ is smallest

anonymous · Answer

$x^2+4$ will be smallest when $x=0$ and largest when $x=-5$
we can do it the stupid calculus way if you like

anonymous · Answer

you have the derivative correct (ugly but correct)

anonymous · Answer

use fraction notation 
$$f'(x)=\frac{2x}{\sqrt[3]{(x^2+4)^2}}$$

anonymous · Answer

Lol ah okay that makes sense so then would there only be one critical point? x=0?

anonymous · Answer

now the critical points are where it is undefined (where the denominator is zero) or where it is zero (where the numerator is zero

anonymous · Answer

right, only at $x=0$

anonymous · Answer

since $x^2+4\geq 4$ the denominator is never zero

anonymous · Answer

So now I have to plug in 0 and the end points into the original equation right.

anonymous · Answer

but i think if you were not in calculus class, the answer would be totally obvious
the cubed root of something is largest when that something is largest right?  and on that interval $x^2+4$ is largest if $x=-5$

anonymous · Answer

yes plug it in to the ORIGINAL  function not the derivative

anonymous · Answer

Is there any way I can check myself on my TI-84 calculator?

anonymous · Answer

if you plug the critical point in to the derivative you get zero since thats how you got it in the first place

anonymous · Answer

calculators are so last century http://www.wolframalpha.com/input/?i=%28x^2%2B4%29^1%2F3

anonymous · Answer

this is even better for your interval http://www.wolframalpha.com/input/?i=%28x^2%2B4%29^1%2F3+domain+-5..4

anonymous · Answer

Well the only reason I'm even asking is so I can check myself on the test and because every time I plug all the numbers into the original equation and get the answers and then try use my calculator to solve also I get different answers. So I thought maybe that I was plugging it into my calculator wrong. The min values I got are (0,1.6) and (4,1.6)