# Find the absolute max and absolute min

• Nov 21st 2009, 10:53 AM
s3a
Find the absolute max and absolute min
Question:
(1 pt) Let http://gauss.vaniercollege.qc.ca/web...bec85871c1.png on the interval http://gauss.vaniercollege.qc.ca/web...b98e26d921.png. Find the absolute maximum and absolute minimum of http://gauss.vaniercollege.qc.ca/web...910e13a9c1.png on this interval.

The absolute max occurs at http://gauss.vaniercollege.qc.ca/web...dfef23ff01.png ?
The absolute min occurs at http://gauss.vaniercollege.qc.ca/web...dfef23ff01.png ?

The absolute max is supposedly x = -1 and I do not know what the absolute min is. I use the two "edges" of the constraint and since when I solved for x I already got what the "edges" of the constraint then I determined the two "edges" to be the values to optimize with but I just do not understand what is going on. In addition, my calculator says (-1)^(2/3) does not exist but algebraically, I determined it to be = 1; who's right. me or the calculator? (My work is attached)

Any help would be greatly appreciated!
• Nov 21st 2009, 12:00 PM
earboth
Quote:

Originally Posted by s3a
Question:
(1 pt) Let http://gauss.vaniercollege.qc.ca/web...bec85871c1.png on the interval http://gauss.vaniercollege.qc.ca/web...b98e26d921.png. Find the absolute maximum and absolute minimum of http://gauss.vaniercollege.qc.ca/web...910e13a9c1.png on this interval.

The absolute max occurs at http://gauss.vaniercollege.qc.ca/web...dfef23ff01.png ?
The absolute min occurs at http://gauss.vaniercollege.qc.ca/web...dfef23ff01.png ?

The absolute max is supposedly x = -1 and I do not know what the absolute min is. I use the two "edges" of the constraint and since when I solved for x I already got what the "edges" of the constraint then I determined the two "edges" to be the values to optimize with but I just do not understand what is going on. In addition, my calculator says (-1)^(2/3) does not exist but algebraically, I determined it to be = 1; who's right. me or the calculator? (My work is attached)

Any help would be greatly appreciated!

1. f(-1) = 5

f(1) = 1

2. You can show that f(0) = 0 and $\displaystyle f(x) \geq 0$ for all $\displaystyle x \in \mathbb{R}$
• Nov 21st 2009, 12:03 PM
s3a
How would you know to plug in x = 0 when you did not obtain x = 0 when solving nor are the "edges" of the constraint 0?
• Nov 21st 2009, 12:19 PM
Raoh
hi(Hi)
$\displaystyle x=1$ the root of your first derivative is either a maximum or a minimum,try to find $\displaystyle f''$ and study its sign at that point.
• Nov 22nd 2009, 01:03 AM
earboth
Quote:

Originally Posted by s3a
How would you know to plug in x = 0 when you did not obtain x = 0 when solving nor are the "edges" of the constraint 0?

I don't know what you mean by that ...?

$\displaystyle 3\sqrt[3]{x^2} - 2x=0~\implies~\sqrt[3]{x^2}(3-2\sqrt[3]{x})=0~\implies~x=0~\vee~x=\left(\frac32 \right)^3$
• Nov 22nd 2009, 03:36 AM
HallsofIvy
Quote:

Originally Posted by s3a
How would you know to plug in x = 0 when you did not obtain x = 0 when solving nor are the "edges" of the constraint 0?

The max or min may occur where the derivative is 0 or where the derivative does not exist. The derivative does not exist at x= 0.
• Nov 22nd 2009, 03:38 AM
HallsofIvy
Quote:

Originally Posted by Raoh
hi(Hi)
$\displaystyle x=1$ the root of your first derivative is either a maximum or a minimum,try to find $\displaystyle f''$ and study its sign at that point.

For absolute max or min, the second derivative test is not necessary. Just calculate the values at the endpoints and critical points (where the derivative is 0 or does not exist) and see which value is highest and which lowest.