# Finding Critical Numbers and Absolute Extrema's.

• October 26th 2008, 10:49 AM
teddybear
Finding Critical Numbers and Absolute Extrema's.
1) ${\frac{3}{2}}x^{\frac{1}{2}}-{\frac{3}{2}}x^{\frac{-1}{2}}$

For this one... I found that my critical number was 1. I don't know if its right.
I was also give the Interval [-2,3]. So I plugged it all into my original function of $f(x)=(x-3)x^{\frac{1}{2}}$

After I plugged it in, I got

f(1)=2
f(3)=0
f(-2)=undefined.

But then again, I don't know if 1 is the critical number. And.. since it's undefined.. I just leave it alone right? So the max would be (1,2) and min would be (3,0) assuming that 1 is the critical number?

2) $g(x) = \frac{-4x+6}{(x^2-3x)^3}$

Found that the critical number was ${\frac{3}{2}}$
Given interval [-1,6]

I need help finding the absolute extremas here even though its just plugging in (Lipssealed)

Suppose to plug it into this original function:
$g(x)=(x^2-3x)^{-2} + 2$
• October 27th 2008, 11:45 AM
Moo
Hello,
Quote:

Originally Posted by teddybear
1) ${\frac{3}{2}}x^{\frac{1}{2}}-{\frac{3}{2}}x^{\frac{-1}{2}}$

For this one... I found that my critical number was 1. I don't know if its right.
I was also give the Interval [-2,3]. So I plugged it all into my original function of $f(x)=(x-3)x^{\frac{1}{2}}$

After I plugged it in, I got

f(1)=2
f(3)=0
f(-2)=undefined.

But then again, I don't know if 1 is the critical number. And.. since it's undefined.. I just leave it alone right? So the max would be (1,2) and min would be (3,0) assuming that 1 is the critical number?

Yep, 1 is the critical number.
There is no reason to include -2 and 3 in the list of critical points. especially when it's not defined :p
The conditions for a point to be critical are :
- derivative = 0. << x=1 is the only solution
- or undefined for the derivative but defined for the function itself. << the domain of the function is $x \geq 0$. At x=0, the derivative is not defined. So you can see for that.

The boundaries can intervene in the absolute extremas list, but not really in the critical points list.

For the extremas, you have to check that any point in the interval is < or > to the extrema.

Quote:

2) $g(x) = \frac{-4x+6}{(x^2-3x)^3}$

Found that the critical number was ${\frac{3}{2}}$
Given interval [-1,6]

I need help finding the absolute extremas here even though its just plugging in (Lipssealed)

Suppose to plug it into this original function:
$g(x)=(x^2-3x)^{-2} + 2$
$g(\tfrac 32)=\frac{166}{81}$ (not sure, but you can check it (Surprised))

See if for any value of x, $(x^2-3x)^{-2}+2\geq \text{ or } \leq \frac{166}{81}$