# Crit points, extreme values, and basic algebra

• Nov 7th 2008, 05:36 AM
Miss Kit'n
Crit points, extreme values, and basic algebra
Consider the function g(x) = x^2sqrt(5-x)

a)find the intervals on which the function is increasing and decreasing.

I'm stuck at basic algebra here and not sure why. This is what I've got.

g'(x) = 2x*sqrt(5-x) - ((x^2)/(2sqrt(5-x)))

Then I set = 0 and solve for x...

2x*sqrt(5-x) - ((x^2)/(2sqrt(5-x))) = 0
(x^2)/(2sqrt(5-x)) = (2x*sqrt(5-x))
x^2 = (2x*sqrt(5-x))/(2sqrt(5-x))

now, I'm trying to simplify by cancelation but can't remember if I can cancel out the two sqrts or the 2's or both?

Thanks!
• Nov 7th 2008, 06:01 AM
ajj86
Response
Consider the function g(x) = x^2sqrt(5-x)

a)find the intervals on which the function is increasing and decreasing.

Well, start by taking g'(x) by using the product rule:
g'(x) = 2xsqrt(5-x) + x^2*(1/2)(5-x)^(-1/2)*-1
g'(x) = 2x*sqrt(5-x) -(1/2)*x^2/sqrt(5-x) = 0
2x*sqrt(5-x) = x^2/(2*sqrt(5-x))

Multiply both sides by 2*sqrt(5-x) to get:
4x*(5-x) = x^2
20x - 4x^2 = x^2
5x^2 - 20x = 0
5x(x-4) = 0

x = 0, x = 4
Hope this helps.
• Nov 7th 2008, 06:08 AM
ajj86
5x(x-4) = 0
x = 0, x = 4

Pick a point to the left of zero, say at x=-1. This causes the 5x factor to be negative, and the (x-4) factor to be negative. Their resulting product is positive, so this interval is increasing.

Any point in between x=0 and x=4, say at x = 2 for example. This causes the 5x factor to be positive and the (x-4) factor to be negative. Their resulting product is negative, so this interval is decreasing.

Any point to the right of x = 4, say at x = 5. This causes the 5x factor to be positive and the (x-4) factor to be positive. Their resulting product is positive, so this interval is increasing.

The final thing looks like:

Increasing: (-infinity, 0) U (4, infinity)
Decreasing: (0,4)
• Nov 7th 2008, 06:38 AM
Miss Kit'n
thanks a lot for taking time for this! Dunno why I keep getting hung up on it :(